《c++程序设计》谭浩强课后习题答案

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(完整版)谭浩强c程序设计课后习题答案

(完整版)谭浩强c程序设计课后习题答案

谭浩强c++程序设计课后答案娄警卫第一章1.5题#include <iostream> using namespace std; int main(){cout<<"This"<<"is"; cout<<"a"<<"C++"; cout<<"program."; return 0;1.6题#include <iostream> using namespace std; int main(){int a,b,c;a=10;b=23;c=a+b;cout<<"a+b="; cout<<c;cout<<endl;return 0;}1.7七题#include <iostream> using namespace std; int main(){int a,b,c;int f(int x,int y,int z); cin>>a>>b>>c;c=f(a,b,c);cout<<c<<endl; return 0;}int f(int x,int y,int z) {int m;if (x<y) m=x;else m=y;if (z<m) m=z;return(m);}1.8题#include <iostream>using namespace std;int main(){int a,b,c;cin>>a>>b;c=a+b;cout<<"a+b="<<a+b<<endl; return 0;}1.9题#include <iostream>using namespace std;int main(){int a,b,c;int add(int x,int y); cin>>a>>b;c=add(a,b);cout<<"a+b="<<c<<endl; return 0;}int add(int x,int y){int z;z=x+y;return(z);}第二章2.3题#include <iostream>using namespace std;int main(){char c1='a',c2='b',c3='c',c4='\101',c5='\116'; cout<<c1<<c2<<c3<<'\n';cout<<"\t\b"<<c4<<'\t'<<c5<<'\n';return 0;}2.4题#include <iostream>using namespace std;int main(){char c1='C',c2='+',c3='+';cout<<"I say: \""<<c1<<c2<<c3<<'\"';cout<<"\t\t"<<"He says: \"C++ is very interesting!\""<< '\n';return 0;}2.7题#include <iostream>using namespace std;int main(){int i,j,m,n;i=8;j=10;m=++i+j++;n=(++i)+(++j)+m;cout<<i<<'\t'<<j<<'\t'<<m<<'\t'<<n<<endl; return 0;}2.8题#include <iostream>using namespace std;int main(){char c1='C', c2='h', c3='i', c4='n', c5='a';c1+=4;c2+=4;c3+=4;c4+=4;c5+=4;cout<<"password is:"<<c1<<c2<<c3<<c4<<c5<<endl;return 0;}第三章3.2题#include <iostream>#include <iomanip>using namespace std;int main ( ){float h,r,l,s,sq,vq,vz;const float pi=3.1415926;cout<<"please enter r,h:";cin>>r>>h;l=2*pi*r;s=r*r*pi;sq=4*pi*r*r;vq=3.0/4.0*pi*r*r*r;vz=pi*r*r*h;cout<<setiosflags(ios::fixed)<<setiosflags(ios:: right)<<setprecision(2);cout<<"l= "<<setw(10)<<l<<endl;cout<<"s= "<<setw(10)<<s<<endl;cout<<"sq="<<setw(10)<<sq<<endl;cout<<"vq="<<setw(10)<<vq<<endl;cout<<"vz="<<setw(10)<<vz<<endl;return 0;}3.3题#include <iostream>using namespace std;int main (){float c,f;cout<<"请输入一个华氏温度:";cin>>f;c=(5.0/9.0)*(f-32); //注意5和9要用实型表示,否则5/9值为0cout<<"摄氏温度为:"<<c<<endl;return 0;};3.4题#include <iostream>using namespace std;int main ( ){char c1,c2;cout<<"请输入两个字符c1,c2:";c1=getchar(); //将输入的第一个字符赋给c1c2=getchar(); //将输入的第二个字符赋给c2cout<<"用putchar函数输出结果为:"; putchar(c1);putchar(c2);cout<<endl;cout<<"用cout语句输出结果为:";cout<<c1<<c2<<endl;return 0;}3.4题另一解#include <iostream>using namespace std;int main ( ){char c1,c2;cout<<"请输入两个字符c1,c2:";c1=getchar(); //将输入的第一个字符赋给c1c2=getchar(); //将输入的第二个字符赋给c2cout<<"用putchar函数输出结果为:"; putchar(c1);putchar(44);putchar(c2);cout<<endl;cout<<"用cout语句输出结果为:";cout<<c1<<","<<c2<<endl;return 0;}3.5题#include <iostream>using namespace std;int main ( ){char c1,c2;int i1,i2; //定义为整型cout<<"请输入两个整数i1,i2:";cin>>i1>>i2;c1=i1;c2=i2;cout<<"按字符输出结果为:"<<c1<<" , "<<c2<<endl;return 0;}3.8题#include <iostream>using namespace std;int main ( ){ int a=3,b=4,c=5,x,y;cout<<(a+b>c && b==c)<<endl;cout<<(a||b+c && b-c)<<endl;cout<<(!(a>b) && !c||1)<<endl;cout<<(!(x=a) && (y=b) && 0)<<endl;cout<<(!(a+b)+c-1 && b+c/2)<<endl; return 0;}3.9题include <iostream>using namespace std;int main ( ){int a,b,c;cout<<"please enter three integer numbers:";cin>>a>>b>>c;if(a<b)if(b<c)cout<<"max="<<c;elsecout<<"max="<<b;else if (a<c)cout<<"max="<<c;elsecout<<"max="<<a;cout<<endl;return 0;}3.9题另一解#include <iostream>using namespace std;int main ( ){int a,b,c,temp,max ;cout<<"please enter three integer numbers:";cin>>a>>b>>c;temp=(a>b)?a:b; /* 将a和b中的大者存入temp中*/max=(temp>c)?temp:c; /* 将a和b中的大者与c比较,最大者存入max*/cout<<"max="<<max<<endl;return 0;}3.10题#include <iostream>using namespace std;int main ( ){int x,y;cout<<"enter x:";cin>>x;if (x<1){y=x;cout<<"x="<<x<<", y=x="<<y;}else if (x<10) // 1≤x<10{y=2*x-1;cout<<"x="<<x<<", y=2*x-1="<<y;}else// x≥10{y=3*x-11;cout<<"x="<<x<<",y=3*x-11="<<y;}cout<<endl;return 0;}3.11题#include <iostream>using namespace std; int main (){float score;char grade;cout<<"please enter score of student:"; cin>>score;while (score>100||score<0){cout<<"data error,enter data again.";cin>>score;}switch(int(score/10)){case 10:case 9: grade='A';break;case 8: grade='B';break;case 7: grade='C';break;case 6: grade='D';break;default:grade='E';}cout<<"score is "<<score<<", grade is "<<grade<<endl;return 0;}3.12题#include <iostream>using namespace std;int main (){long int num;intindiv,ten,hundred,thousand,ten_thousand,pla ce;/*分别代表个位,十位,百位,千位,万位和位数*/cout<<"enter an integer(0~99999):"; cin>>num;if (num>9999)place=5;else if (num>999)place=4;else if (num>99)place=3;else if (num>9)place=2;else place=1;cout<<"place="<<place<<endl;//计算各位数字ten_thousand=num/10000;thousand=(int)(num-ten_thousand*10000)/1 000;hundred=(int)(num-ten_thousand*10000-tho usand*1000)/100;ten=(int)(num-ten_thousand*10000-thousan d*1000-hundred*100)/10;indiv=(int)(num-ten_thousand*10000-thousa nd*1000-hundred*100-ten*10);cout<<"original order:";switch(place){case5:cout<<ten_thousand<<","<<thousand<<","< <hundred<<","<<ten<<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<thousand<<ten _thousand<<endl;break;case4:cout<<thousand<<","<<hundred<<","<<ten <<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<thousand<<en dl;break;case3:cout<<hundred<<","<<ten<<","<<indiv<<en dl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<endl;break;case 2:cout<<ten<<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<endl;break;case 1:cout<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<endl;break;}return 0;}3.13题#include <iostream>using namespace std;int main (){ long i; //i为利润floatbonus,bon1,bon2,bon4,bon6,bon10;bon1=100000*0.1; //利润为10万元时的奖金bon2=bon1+100000*0.075; //利润为20万元时的奖金bon4=bon2+100000*0.05; //利润为40万元时的奖金bon6=bon4+100000*0.03; //利润为60万元时的奖金bon10=bon6+400000*0.015; //利润为100万元时的奖金cout<<"enter i:";cin>>i;if (i<=100000)bonus=i*0.1;//利润在10万元以内按10%提成奖金else if (i<=200000)bonus=bon1+(i-100000)*0.075; //利润在10万元至20万时的奖金else if (i<=400000)bonus=bon2+(i-200000)*0.05; //利润在20万元至40万时的奖金else if (i<=600000)bonus=bon4+(i-400000)*0.03; //利润在40万元至60万时的奖金else if (i<=1000000)bonus=bon6+(i-600000)*0.015; //利润在60万元至100万时的奖金elsebonus=bon10+(i-1000000)*0.01; //利润在100万元以上时的奖金cout<<"bonus="<<bonus<<endl;return 0;}3.13题另一解#include <iostream>using namespace std;int main (){long i;float bonus,bon1,bon2,bon4,bon6,bon10; int c;bon1=100000*0.1;bon2=bon1+100000*0.075;bon4=bon2+200000*0.05;bon6=bon4+200000*0.03;bon10=bon6+400000*0.015;cout<<"enter i:";cin>>i;c=i/100000;if (c>10) c=10;switch(c){case 0: bonus=i*0.1; break;case 1: bonus=bon1+(i-100000)*0.075; break;case 2:case3:bonus=bon2+(i-200000)*0.05;break;case 4:case5:bonus=bon4+(i-400000)*0.03;break;case 6:case 7:case 8:case 9: bonus=bon6+(i-600000)*0.015; break;case 10: bonus=bon10+(i-1000000)*0.01;}cout<<"bonus="<<bonus<<endl;return 0;}3.14题#include <iostream>using namespace std;int main (){int t,a,b,c,d;cout<<"enter four numbers:";cin>>a>>b>>c>>d;cout<<"a="<<a<<", b="<<b<<", c="<<c<<",d="<<d<<endl;if (a>b){t=a;a=b;b=t;}if (a>c){t=a; a=c; c=t;}if (a>d){t=a; a=d; d=t;}if (b>c){t=b; b=c; c=t;}if (b>d){t=b; b=d; d=t;}if (c>d){t=c; c=d; d=t;}cout<<"the sorted sequence:"<<endl;cout<<a<<", "<<b<<", "<<c<<", "<<d<<endl; return 0;}3.15题#include <iostream>using namespace std;int main (){int p,r,n,m,temp;cout<<"please enter two positive integer numbers n,m:";cin>>n>>m;if (n<m){temp=n;n=m;m=temp; //把大数放在n中, 小数放在m中}p=n*m; //先将n和m的乘积保存在p中, 以便求最小公倍数时用while (m!=0) //求n和m的最大公约数{r=n%m;n=m;m=r;}cout<<"HCF="<<n<<endl;cout<<"LCD="<<p/n<<endl; // p是原来两个整数的乘积return 0;}3.16题#include <iostream>using namespace std;int main (){char c;int letters=0,space=0,digit=0,other=0;cout<<"enter one line::"<<endl;while((c=getchar())!='\n'){if (c>='a' && c<='z'||c>='A' && c<='Z')letters++;else if (c==' ')space++;else if (c>='0' && c<='9')digit++;elseother++;}cout<<"letter:"<<letters<<", space:"<<space<<", digit:"<<digit<<", other:"<<other<<endl;return 0;}3.17题#include <iostream>using namespace std;int main (){int a,n,i=1,sn=0,tn=0;cout<<"a,n=:";cin>>a>>n;while (i<=n){tn=tn+a; //赋值后的tn为i个a 组成数的值sn=sn+tn; //赋值后的sn为多项式前i项之和a=a*10;++i;}cout<<"a+aa+aaa+...="<<sn<<endl;return 0;}3.18题#include <iostream>using namespace std;int main (){float s=0,t=1;int n;for (n=1;n<=20;n++){t=t*n; // 求n!s=s+t; // 将各项累加}cout<<"1!+2!+...+20!="<<s<<endl;return 0;}3.19题#include <iostream>using namespace std;int main (){int i,j,k,n;cout<<"narcissus numbers are:"<<endl;for (n=100;n<1000;n++){i=n/100;j=n/10-i*10;k=n%10;if (n == i*i*i + j*j*j + k*k*k)cout<<n<<" ";}cout<<endl;return 0;}3.20题#include <iostream>using namespace std;int main(){const int m=1000; // 定义寻找范围int k1,k2,k3,k4,k5,k6,k7,k8,k9,k10;int i,a,n,s;for (a=2;a<=m;a++) // a是2~1000之间的整数,检查它是否为完数{n=0; // n用来累计a的因子的个数s=a; // s用来存放尚未求出的因子之和,开始时等于afor (i=1;i<a;i++) // 检查i是否为a 的因子if (a%i==0) // 如果i是a的因子{n++; // n加1,表示新找到一个因子s=s-i; // s减去已找到的因子,s的新值是尚未求出的因子之和switch(n) // 将找到的因子赋给k1,...,k10{case 1:k1=i; break; // 找出的笫1个因子赋给k1case 2:k2=i; break; // 找出的笫2个因子赋给k2case 3:k3=i; break; // 找出的笫3个因子赋给k3case 4:k4=i; break; // 找出的笫4个因子赋给k4case 5:k5=i; break; // 找出的笫5个因子赋给k5case 6:k6=i; break; // 找出的笫6个因子赋给k6case 7:k7=i; break; // 找出的笫7个因子赋给k7case 8:k8=i; break; // 找出的笫8个因子赋给k8case 9:k9=i; break; // 找出的笫9个因子赋给k9case 10:k10=i; break; // 找出的笫10个因子赋给k10}}if (s==0) // s=0表示全部因子都已找到了{cout<<a<<" is a 完数"<<endl;cout<<"its factors are:";if (n>1) cout<<k1<<","<<k2; // n>1表示a至少有2个因子if (n>2) cout<<","<<k3; // n>2表示至少有3个因子,故应再输出一个因子if (n>3) cout<<","<<k4; // n>3表示至少有4个因子,故应再输出一个因子if (n>4) cout<<","<<k5; // 以下类似if (n>5) cout<<","<<k6;if (n>6) cout<<","<<k7;if (n>7) cout<<","<<k8;if (n>8) cout<<","<<k9;if (n>9) cout<<","<<k10;cout<<endl<<endl;}}return 0;}3.20题另一解#include <iostream>using namespace std;int main(){int m,s,i;for (m=2;m<1000;m++){s=0;for (i=1;i<m;i++)if ((m%i)==0) s=s+i;if(s==m){cout<<m<<" is a完数"<<endl;cout<<"its factors are:";for (i=1;i<m;i++)if (m%i==0) cout<<i<<" ";cout<<endl;}}return 0;}3.20题另一解#include <iostream>using namespace std;int main(){int k[11];int i,a,n,s;for (a=2;a<=1000;a++){n=0;s=a;for (i=1;i<a;i++)if ((a%i)==0){n++;s=s-i;k[n]=i; // 将找到的因子赋给k[1]┅k[10]}if (s==0){cout<<a<<" is a 完数"<<endl;cout<<"its factors are:";for (i=1;i<n;i++)cout<<k[i]<<" ";cout<<k[n]<<endl;}}return 0;}3.21题#include <iostream>using namespace std;int main(){int i,t,n=20;double a=2,b=1,s=0;for (i=1;i<=n;i++){s=s+a/b;t=a;a=a+b; // 将前一项分子与分母之和作为下一项的分子b=t; // 将前一项的分子作为下一项的分母}cout<<"sum="<<s<<endl;return 0;} 3.22题#include <iostream>using namespace std;int main(){int day,x1,x2;day=9;x2=1;while(day>0){x1=(x2+1)*2; // 第1天的桃子数是第2天桃子数加1后的2倍x2=x1;day--;}cout<<"total="<<x1<<endl;return 0;}3.23题#include <iostream>#include <cmath>using namespace std;int main(){float a,x0,x1;cout<<"enter a positive number:"; cin>>a; // 输入a的值x0=a/2;x1=(x0+a/x0)/2;do{x0=x1;x1=(x0+a/x0)/2;}while(fabs(x0-x1)>=1e-5);cout<<"The square root of "<<a<<" is "<<x1<<endl;return 0;}3.24题#include <iostream>using namespace std;int main(){int i,k;for (i=0;i<=3;i++) // 输出上面4行*号{for (k=0;k<=2*i;k++)cout<<"*"; // 输出*号cout<<endl; //输出完一行*号后换行}for (i=0;i<=2;i++) // 输出下面3行*号{for (k=0;k<=4-2*i;k++)cout<<"*"; // 输出*号cout<<endl; // 输出完一行*号后换行}return 0;}3.25题#include <iostream>using namespace std;int main(){char i,j,k; /* i是a的对手;j是b的对手;k是c的对手*/for (i='X';i<='Z';i++)for (j='X';j<='Z';j++)if (i!=j)for (k='X';k<='Z';k++)if (i!=k && j!=k)if (i!='X' && k!='X' && k!='Z')cout<<"A--"<<i<<"B--"<<j<<" C--"<<k<<endl;return 0;}第四章4.1题#include <iostream>using namespace std;int main(){int hcf(int,int);int lcd(int,int,int);int u,v,h,l;cin>>u>>v;h=hcf(u,v);cout<<"H.C.F="<<h<<endl;l=lcd(u,v,h);cout<<"L.C.D="<<l<<endl;return 0;}int hcf(int u,int v){int t,r;if (v>u){t=u;u=v;v=t;}while ((r=u%v)!=0){u=v;v=r;}return(v);}int lcd(int u,int v,int h){return(u*v/h);}4.2题#include <iostream>#include <math.h>using namespace std;float x1,x2,disc,p,q;int main(){void greater_than_zero(float,float); void equal_to_zero(float,float);void smaller_than_zero(float,float); float a,b,c;cout<<"input a,b,c:";cin>>a>>b>>c;disc=b*b-4*a*c;cout<<"root:"<<endl;if (disc>0){greater_than_zero(a,b);cout<<"x1="<<x1<<",x2="<<x2<<endl; }else if (disc==0){equal_to_zero(a,b);cout<<"x1="<<x1<<",x2="<<x2<<endl;}else{smaller_than_zero(a,b);cout<<"x1="<<p<<"+"<<q<<"i"<<endl;cout<<"x2="<<p<<"-"<<q<<"i"<<endl;}return 0;}void greater_than_zero(float a,float b) /* 定义一个函数,用来求disc>0时方程的根*/{x1=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);}void equal_to_zero(float a,float b) /* 定义一个函数,用来求disc=0时方程的根*/{x1=x2=(-b)/(2*a);}void smaller_than_zero(float a,float b) /* 定义一个函数,用来求disc<0时方程的根*/{p=-b/(2*a);q=sqrt(-disc)/(2*a);}4.3题#include <iostream>using namespace std;int main(){int prime(int); /* 函数原型声明*/int n;cout<<"input an integer:";cin>>n;if (prime(n))cout<<n<<" is a prime."<<endl;elsecout<<n<<" is not a prime."<<endl;return 0;}int prime(int n){int flag=1,i;for (i=2;i<n/2 && flag==1;i++)if (n%i==0)flag=0;return(flag);}4.4题#include <iostream>using namespace std;int main(){int fac(int);int a,b,c,sum=0;cout<<"enter a,b,c:";cin>>a>>b>>c;sum=sum+fac(a)+fac(b)+fac(c);cout<<a<<"!+"<<b<<"!+"<<c<<"!="<<sum<<e ndl;return 0;}int fac(int n){int f=1;for (int i=1;i<=n;i++)f=f*i;return f;}4.5题#include <iostream>#include <cmath>using namespace std;int main(){double e(double);double x,sinh;cout<<"enter x:";cin>>x;sinh=(e(x)+e(-x))/2;cout<<"sinh("<<x<<")="<<sinh<<endl;return 0;}double e(double x){return exp(x);}4.6题#include <iostream>#include <cmath>using namespace std;int main(){doublesolut(double ,double ,double ,double ); double a,b,c,d;cout<<"input a,b,c,d:";cin>>a>>b>>c>>d;cout<<"x="<<solut(a,b,c,d)<<endl;return 0;}double solut(double a,double b,double c,double d){double x=1,x0,f,f1;do{x0=x;f=((a*x0+b)*x0+c)*x0+d;f1=(3*a*x0+2*b)*x0+c;x=x0-f/f1;}while(fabs(x-x0)>=1e-5);return(x);}4.7题#include <iostream>#include <cmath>using namespace std;int main(){void godbaha(int);int n;cout<<"input n:";cin>>n;godbaha(n);return 0;}void godbaha(int n){int prime(int);int a,b;for(a=3;a<=n/2;a=a+2){if(prime(a)){b=n-a;if (prime(b))cout<<n<<"="<<a<<"+"<<b<<endl;}}}int prime(int m){int i,k=sqrt(m);for(i=2;i<=k;i++)if(m%i==0) break;if (i>k) return 1;else return 0;}4.8题#include <iostream>using namespace std;int main(){int x,n;float p(int,int);cout<<"input n & x:";cin>>n>>x;cout<<"n="<<n<<",x="<<x<<endl;;cout<<"P"<<n<<"(x)="<<p(n,x)<<endl; return 0;}float p(int n,int x){if (n==0)return(1);else if (n==1)return(x);elsereturn(((2*n-1)*x*p((n-1),x)-(n-1)*p((n-2),x))/n);}4.9题#include <iostream>using namespace std;int main(){void hanoi(int n,char one,char two,char three);int m;cout<<"input the number of diskes:"; cin>>m;cout<<"The steps of moving "<<m<<" disks:"<<endl;hanoi(m,'A','B','C');return 0;}void hanoi(int n,char one,char two,char three) //将n个盘从one座借助two座,移到three座{void move(char x,char y);if(n==1) move(one,three);else{hanoi(n-1,one,three,two);move(one,three);hanoi(n-1,two,one,three);}}void move(char x,char y){cout<<x<<"-->"<<y<<endl;}4.10题#include <iostream>using namespace std;int main(){void convert(int n);int number;cout<<"input an integer:";cin>>number;cout<<"output:"<<endl;if (number<0){cout<<"-";number=-number;}convert(number);cout<<endl;return 0;}void convert(int n){int i;char c;if ((i=n/10)!=0)convert(i);c=n%10+'0';cout<<" "<<c;}4.11题#include <iostream>using namespace std;int main(){int f(int);int n,s;cout<<"input the number n:";cin>>n;s=f(n);cout<<"The result is "<<s<<endl;return 0;}int f(int n){;if (n==1)return 1;elsereturn (n*n+f(n-1));}4.12题#include <iostream>#include <cmath>using namespace std;#define S(a,b,c) (a+b+c)/2#define AREA(a,b,c)sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(S(a,b,c) -c))int main(){float a,b,c;cout<<"input a,b,c:";cin>>a>>b>>c;if (a+b>c && a+c>b && b+c>a)cout<<"area="<<AREA(a,b,c)<<endl; elsecout<<"It is not a triangle!"<<endl; return 0;}4.14题#include <iostream>using namespace std;//#define LETTER 1int main(){char c;cin>>c;#if LETTERif(c>='a' && c<='z')c=c-32;#elseif(c>='A' && c<='Z')c=c+32;#endifcout<<c<<endl;return 0;}4.15题#include <iostream>using namespace std;#define CHANGE 1int main(){char ch[40];cout<<"input text:"<<endl;;gets(ch);#if (CHANGE){for (int i=0;i<40;i++){if (ch[i]!='\0')if (ch[i]>='a'&& ch[i]<'z'||ch[i]>'A'&& ch[i]<'Z')ch[i]+=1;else if (ch[i]=='z'||ch[i]=='Z')ch[i]-=25;}}#endifcout<<"output:"<<endl<<ch<<endl;return 0;}4.16题file#include <iostream>using namespace std;int a;int main(){extern int power(int);int b=3,c,d,m;cout<<"enter an integer a and its power m:"<<endl;cin>>a>>m;c=a*b;cout<<a<<"*"<<b<<"="<<c<<endl;d=power(m);cout<<a<<"**"<<m<<"="<<d<<endl; return 0;}4.16题fileextern int a;int power(int n){int i,y=1;for(i=1;i<=n;i++)y*=a;return y;}第五章5.1题#include <iostream>#include <iomanip>using namespace std;#include <math.h>int main(){int i,j,n,a[101];for (i=1;i<=100;i++)a[i]=i;a[1]=0;for (i=2;i<sqrt(100);i++)for (j=i+1;j<=100;j++){if(a[i]!=0 && a[j]!=0)if (a[j]%a[i]==0)a[j]=0; }cout<<endl;for (i=1,n=0;i<=100;i++){if (a[i]!=0){cout<<setw(5)<<a[i]<<" ";n++;}if(n==10){cout<<endl;n=0;}}cout<<endl;return 0;}5.2题#include <iostream>using namespace std;//#include <math.h>int main(){int i,j,min,temp,a[11];cout<<"enter data:"<<endl;for (i=1;i<=10;i++){cout<<"a["<<i<<"]=";cin>>a[i]; //输入10个数}cout<<endl<<"The original numbers:"<<endl;;for (i=1;i<=10;i++)cout<<a[i]<<" "; // 输出这10个数cout<<endl;;for (i=1;i<=9;i++) //以下8行是对10个数排序{min=i;for (j=i+1;j<=10;j++)if (a[min]>a[j]) min=j;temp=a[i]; //以下3行将a[i+1]~a[10]中最小者与a[i] 对换a[i]=a[min];a[min]=temp;}cout<<endl<<"The sorted numbers:"<<endl;for (i=1;i<=10;i++) // 输出已排好序的10个数cout<<a[i]<<" ";cout<<endl;return 0;}5.3题#include <iostream>using namespace std;int main(){int a[3][3],sum=0;int i,j;cout<<"enter data:"<<endl;;for (i=0;i<3;i++)for (j=0;j<3;j++)cin>>a[i][j];for (i=0;i<3;i++)sum=sum+a[i][i];cout<<"sum="<<sum<<endl;return 0;}5.4题#include <iostream>using namespace std;int main(){int a[11]={1,4,6,9,13,16,19,28,40,100};int num,i,j;cout<<"array a:"<<endl;for (i=0;i<10;i++)cout<<a[i]<<" ";cout<<endl;;cout<<"insert data:";cin>>num;if (num>a[9])a[10]=num;else。

C程序设计(谭浩强)课后题答案

C程序设计(谭浩强)课后题答案

3If;C程序设计(第三版)[谭浩强著]课后题答案!!!1.5请参照本章例题,编写一个C程序,输出以下信息:**************************Very Good!**************************解:#include “stdio.h”void main(){printf(“**************************”);printf(“\n”);printf(“Very Good!\n”);prin tf(“\n”);printf(“**************************\n”);}1.6 编写一个程序,输入a、b、c三个值,输出其中最大值。

解:#include “stdio.h”void main(){int a,b,c,max;printf(“请输入三个数a,b,c:\n”);scanf(“%d,%d,%d”,&a,&b,&c);max=a;if(max<b)max=b;if(max<c)max=c;printf(“最大数为:%d\n”,max);}第三章3.6写出以下程序运行的结果。

#include “stdio.h”void main(){char c1=’a’,c2=’b’,c3=’c’,c4=’\101’,c5=’\116’;printf(“a%c b%c\tc%c\tabc\n”,c1,c2,c3);printf(“\t\b%c %c”,c4,c5);}解:aaㄩbbㄩㄩㄩccㄩㄩㄩㄩㄩㄩabcAㄩN3.7 要将"China"译成密码,译码规律是:用原来字母后面的第4个字母代替原来的字母.例如,字母"A"后面第4个字母是"E"."E"代替"A"。

因此,"China"应译为"Glmre"。

C语言程序设计第三版谭浩强课后习题答案完整版

C语言程序设计第三版谭浩强课后习题答案完整版
main()
{
int a,b;float x,y;char c1c2;
scanf("a=%d_b=%d," &a,&b);
scanf("_x=%f_y=%e",&x,&y);
scanf("_c1 =%c_c2=%c," &c1,&c2);
}
a=3_b=7
_x=8.5_y=71.82
_c1=A_c2=a
1
main()
scanf("%d,%d,%d",&a,&b,&c);
{int a,b,c;
temp=(a>b)?a:b;/*
将a和b中
printf("请输入3个整数:");
的大者存人temp中*/
scanf("%d,%d,%d",&a,&b,&c);
max=(temp>c)?temp:c;/*的大者与c比较,取最大者*/
(4)!(x=a)&&(y=b)&&0
(5)!(a+b)+c-1&&b+c/2
解:
(1) 0
(2)1
(3)1
(4)0
(5)1
解:设有一个逻辑表达式, 若其结果为“真”,
5.4有3个整数a b、c,由键盘输入,
方法一
{int a,b,c,temp,max;
#include <stdio.h>
printf("请输入3个整数:");
3.12写出下面表达式运算后a的值,设原

《C语言程序设计》课后习题答案()谭浩强

《C语言程序设计》课后习题答案()谭浩强

第1章程序设计和C语言11.1什么是计算机程序11.2什么是计算机语言11.3C语言的发展及其特点31.4最简单的C语言程序51.4.1最简单的C语言程序举例61.4.2C语言程序的结构101.5运行C程序的步骤与方法121.6程序设计的任务141-5 #include <stdio.h>int main ( ){ printf ("**************************\n\n");printf(" Very Good!\n\n");printf ("**************************\n");return 0;}1-6#include <stdio.h>int main(){int a,b,c,max;printf("please input a,b,c:\n");scanf("%d,%d,%d",&a,&b,&c);max=a;if (max<b)max=b;if (max<c)max=c;printf("The largest number is %d\n",max); return 0;}第2章算法——程序的灵魂162.1什么是算法162.2简单的算法举例172.3算法的特性212.4怎样表示一个算法222.4.1用自然语言表示算法222.4.2用流程图表示算法222.4.3三种基本结构和改进的流程图262.4.4用N S流程图表示算法282.4.5用伪代码表示算法312.4.6用计算机语言表示算法322.5结构化程序设计方法34习题36第章最简单的C程序设计——顺序程序设计37 3.1顺序程序设计举例373.2数据的表现形式及其运算393.2.1常量和变量393.2.2数据类型423.2.3整型数据443.2.4字符型数据473.2.5浮点型数据493.2.6怎样确定常量的类型513.2.7运算符和表达式523.3C语句573.3.1C语句的作用和分类573.3.2最基本的语句——赋值语句593.4数据的输入输出653.4.1输入输出举例653.4.2有关数据输入输出的概念673.4.3用printf函数输出数据683.4.4用scanf函数输入数据753.4.5字符数据的输入输出78习题823-1 #include <stdio.h>#include <math.h>int main(){float p,r,n;r=0.1;n=10;p=pow(1+r,n);printf("p=%f\n",p);return 0;}3-2-1#include <stdio.h>#include <math.h>int main(){float r5,r3,r2,r1,r0,p,p1,p2,p3,p4,p5;p=1000;r5=0.0585;r3=0.054;r2=0.0468;r1=0.0414;r0=0.0072;p1=p*((1+r5)*5); // 一次存5年期p2=p*(1+2*r2)*(1+3*r3); // 先存2年期,到期后将本息再存3年期p3=p*(1+3*r3)*(1+2*r2); // 先存3年期,到期后将本息再存2年期p4=p*pow(1+r1,5); // 存1年期,到期后将本息存再存1年期,连续存5次p5=p*pow(1+r0/4,4*5); // 存活期存款。

《c 程序设计》谭浩强课后习题与答案

《c  程序设计》谭浩强课后习题与答案
using namespace std;
int main ( )
{int x,y;
cout<<"enter x:";
cin>>x;
if (x<1)
{y=x;
cout<<"x="<<x<<", y=x="<<y;
}
else if (x<10) // 1≤x<10
{y=2*x-1;
cout<<"x="<<x<<", y=2*x-1="<<y;
第一章
1.5题
#include <iostream>
using namespace std;
int main()
{
cout<<"This"<<"is";
cout<<"a"<<"C++";
cout<<"program.";
return 0;
1.6题
#include <iostream>
using namespace std;
cin>>a>>b>>c;
temp=(a>b)?a:b; /*将a和b中的大者存入temp中*/
max=(temp>c)?temp:c; /*将a和b中的大者与c比较,最大者存入max
*/
cout<<"max="<<max<<endl;

《C语言程序设计》课后习题答案(第四版)谭浩强

《C语言程序设计》课后习题答案(第四版)谭浩强

第1章程序设计和C语言11.1什么是计算机程序11.2什么是计算机语言11.3C语言的发展及其特点31.4最简单的C语言程序51.4.1最简单的C语言程序举例61.4.2C语言程序的结构101.5运行C程序的步骤与方法121.6程序设计的任务141-5 #include <stdio.h>int main ( ){ printf ("**************************\n\n"); printf(" Very Good!\n\n");printf ("**************************\n"); return 0;}1-6#include <stdio.h>int main(){int a,b,c,max;printf("please input a,b,c:\n");scanf("%d,%d,%d",&a,&b,&c);max=a;if (max<b)max=b;if (max<c)max=c;printf("The largest number is %d\n",max);return 0;}第2章算法——程序的灵魂162.1什么是算法162.2简单的算法举例172.3算法的特性212.4怎样表示一个算法222.4.1用自然语言表示算法222.4.2用流程图表示算法222.4.3三种基本结构和改进的流程图262.4.4用N S流程图表示算法282.4.5用伪代码表示算法312.4.6用计算机语言表示算法322.5结构化程序设计方法34 习题36第章最简单的C程序设计——顺序程序设计37 3.1顺序程序设计举例373.2数据的表现形式及其运算393.2.1常量和变量393.2.2数据类型423.2.3整型数据443.2.4字符型数据473.2.5浮点型数据493.2.6怎样确定常量的类型513.2.7运算符和表达式523.3C语句573.3.1C语句的作用和分类573.3.2最基本的语句——赋值语句593.4数据的输入输出653.4.1输入输出举例653.4.2有关数据输入输出的概念673.4.3用printf函数输出数据683.4.4用scanf函数输入数据753.4.5字符数据的输入输出78习题823-1 #include <stdio.h>#include <math.h>int main(){float p,r,n;r=0.1;n=10;p=pow(1+r,n);printf("p=%f\n",p);return 0;}3-2-1#include <stdio.h>#include <math.h>int main(){float r5,r3,r2,r1,r0,p,p1,p2,p3,p4,p5;p=1000;r5=0.0585;r3=0.054;r2=0.0468;r1=0.0414;r0=0.0072;p1=p*((1+r5)*5); // 一次存5年期p2=p*(1+2*r2)*(1+3*r3); // 先存2年期,到期后将本息再存3年期p3=p*(1+3*r3)*(1+2*r2); // 先存3年期,到期后将本息再存2年期p4=p*pow(1+r1,5); // 存1年期,到期后将本息存再存1年期,连续存5次 p5=p*pow(1+r0/4,4*5); // 存活期存款。

(完整版)谭浩强c程序设计课后习题答案

(完整版)谭浩强c程序设计课后习题答案

谭浩强c++程序设计课后答案娄警卫第一章1.5题#include <iostream> using namespace std; int main(){cout<<"This"<<"is"; cout<<"a"<<"C++"; cout<<"program."; return 0;1.6题#include <iostream> using namespace std; int main(){int a,b,c;a=10;b=23;c=a+b;cout<<"a+b="; cout<<c;cout<<endl;return 0;}1.7七题#include <iostream> using namespace std; int main(){int a,b,c;int f(int x,int y,int z); cin>>a>>b>>c;c=f(a,b,c);cout<<c<<endl; return 0;}int f(int x,int y,int z) {int m;if (x<y) m=x;else m=y;if (z<m) m=z;return(m);}1.8题#include <iostream>using namespace std;int main(){int a,b,c;cin>>a>>b;c=a+b;cout<<"a+b="<<a+b<<endl; return 0;}1.9题#include <iostream>using namespace std;int main(){int a,b,c;int add(int x,int y); cin>>a>>b;c=add(a,b);cout<<"a+b="<<c<<endl; return 0;}int add(int x,int y){int z;z=x+y;return(z);}第二章2.3题#include <iostream>using namespace std;int main(){char c1='a',c2='b',c3='c',c4='\101',c5='\116'; cout<<c1<<c2<<c3<<'\n';cout<<"\t\b"<<c4<<'\t'<<c5<<'\n';return 0;}2.4题#include <iostream>using namespace std;int main(){char c1='C',c2='+',c3='+';cout<<"I say: \""<<c1<<c2<<c3<<'\"';cout<<"\t\t"<<"He says: \"C++ is very interesting!\""<< '\n';return 0;}2.7题#include <iostream>using namespace std;int main(){int i,j,m,n;i=8;j=10;m=++i+j++;n=(++i)+(++j)+m;cout<<i<<'\t'<<j<<'\t'<<m<<'\t'<<n<<endl; return 0;}2.8题#include <iostream>using namespace std;int main(){char c1='C', c2='h', c3='i', c4='n', c5='a';c1+=4;c2+=4;c3+=4;c4+=4;c5+=4;cout<<"password is:"<<c1<<c2<<c3<<c4<<c5<<endl;return 0;}第三章3.2题#include <iostream>#include <iomanip>using namespace std;int main ( ){float h,r,l,s,sq,vq,vz;const float pi=3.1415926;cout<<"please enter r,h:";cin>>r>>h;l=2*pi*r;s=r*r*pi;sq=4*pi*r*r;vq=3.0/4.0*pi*r*r*r;vz=pi*r*r*h;cout<<setiosflags(ios::fixed)<<setiosflags(ios:: right)<<setprecision(2);cout<<"l= "<<setw(10)<<l<<endl;cout<<"s= "<<setw(10)<<s<<endl;cout<<"sq="<<setw(10)<<sq<<endl;cout<<"vq="<<setw(10)<<vq<<endl;cout<<"vz="<<setw(10)<<vz<<endl;return 0;}3.3题#include <iostream>using namespace std;int main (){float c,f;cout<<"请输入一个华氏温度:";cin>>f;c=(5.0/9.0)*(f-32); //注意5和9要用实型表示,否则5/9值为0cout<<"摄氏温度为:"<<c<<endl;return 0;};3.4题#include <iostream>using namespace std;int main ( ){char c1,c2;cout<<"请输入两个字符c1,c2:";c1=getchar(); //将输入的第一个字符赋给c1c2=getchar(); //将输入的第二个字符赋给c2cout<<"用putchar函数输出结果为:"; putchar(c1);putchar(c2);cout<<endl;cout<<"用cout语句输出结果为:";cout<<c1<<c2<<endl;return 0;}3.4题另一解#include <iostream>using namespace std;int main ( ){char c1,c2;cout<<"请输入两个字符c1,c2:";c1=getchar(); //将输入的第一个字符赋给c1c2=getchar(); //将输入的第二个字符赋给c2cout<<"用putchar函数输出结果为:"; putchar(c1);putchar(44);putchar(c2);cout<<endl;cout<<"用cout语句输出结果为:";cout<<c1<<","<<c2<<endl;return 0;}3.5题#include <iostream>using namespace std;int main ( ){char c1,c2;int i1,i2; //定义为整型cout<<"请输入两个整数i1,i2:";cin>>i1>>i2;c1=i1;c2=i2;cout<<"按字符输出结果为:"<<c1<<" , "<<c2<<endl;return 0;}3.8题#include <iostream>using namespace std;int main ( ){ int a=3,b=4,c=5,x,y;cout<<(a+b>c && b==c)<<endl;cout<<(a||b+c && b-c)<<endl;cout<<(!(a>b) && !c||1)<<endl;cout<<(!(x=a) && (y=b) && 0)<<endl;cout<<(!(a+b)+c-1 && b+c/2)<<endl; return 0;}3.9题include <iostream>using namespace std;int main ( ){int a,b,c;cout<<"please enter three integer numbers:";cin>>a>>b>>c;if(a<b)if(b<c)cout<<"max="<<c;elsecout<<"max="<<b;else if (a<c)cout<<"max="<<c;elsecout<<"max="<<a;cout<<endl;return 0;}3.9题另一解#include <iostream>using namespace std;int main ( ){int a,b,c,temp,max ;cout<<"please enter three integer numbers:";cin>>a>>b>>c;temp=(a>b)?a:b; /* 将a和b中的大者存入temp中*/max=(temp>c)?temp:c; /* 将a和b中的大者与c比较,最大者存入max*/cout<<"max="<<max<<endl;return 0;}3.10题#include <iostream>using namespace std;int main ( ){int x,y;cout<<"enter x:";cin>>x;if (x<1){y=x;cout<<"x="<<x<<", y=x="<<y;}else if (x<10) // 1≤x<10{y=2*x-1;cout<<"x="<<x<<", y=2*x-1="<<y;}else// x≥10{y=3*x-11;cout<<"x="<<x<<",y=3*x-11="<<y;}cout<<endl;return 0;}3.11题#include <iostream>using namespace std; int main (){float score;char grade;cout<<"please enter score of student:"; cin>>score;while (score>100||score<0){cout<<"data error,enter data again.";cin>>score;}switch(int(score/10)){case 10:case 9: grade='A';break;case 8: grade='B';break;case 7: grade='C';break;case 6: grade='D';break;default:grade='E';}cout<<"score is "<<score<<", grade is "<<grade<<endl;return 0;}3.12题#include <iostream>using namespace std;int main (){long int num;intindiv,ten,hundred,thousand,ten_thousand,pla ce;/*分别代表个位,十位,百位,千位,万位和位数*/cout<<"enter an integer(0~99999):"; cin>>num;if (num>9999)place=5;else if (num>999)place=4;else if (num>99)place=3;else if (num>9)place=2;else place=1;cout<<"place="<<place<<endl;//计算各位数字ten_thousand=num/10000;thousand=(int)(num-ten_thousand*10000)/1 000;hundred=(int)(num-ten_thousand*10000-tho usand*1000)/100;ten=(int)(num-ten_thousand*10000-thousan d*1000-hundred*100)/10;indiv=(int)(num-ten_thousand*10000-thousa nd*1000-hundred*100-ten*10);cout<<"original order:";switch(place){case5:cout<<ten_thousand<<","<<thousand<<","< <hundred<<","<<ten<<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<thousand<<ten _thousand<<endl;break;case4:cout<<thousand<<","<<hundred<<","<<ten <<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<thousand<<en dl;break;case3:cout<<hundred<<","<<ten<<","<<indiv<<en dl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<endl;break;case 2:cout<<ten<<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<endl;break;case 1:cout<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<endl;break;}return 0;}3.13题#include <iostream>using namespace std;int main (){ long i; //i为利润floatbonus,bon1,bon2,bon4,bon6,bon10;bon1=100000*0.1; //利润为10万元时的奖金bon2=bon1+100000*0.075; //利润为20万元时的奖金bon4=bon2+100000*0.05; //利润为40万元时的奖金bon6=bon4+100000*0.03; //利润为60万元时的奖金bon10=bon6+400000*0.015; //利润为100万元时的奖金cout<<"enter i:";cin>>i;if (i<=100000)bonus=i*0.1;//利润在10万元以内按10%提成奖金else if (i<=200000)bonus=bon1+(i-100000)*0.075; //利润在10万元至20万时的奖金else if (i<=400000)bonus=bon2+(i-200000)*0.05; //利润在20万元至40万时的奖金else if (i<=600000)bonus=bon4+(i-400000)*0.03; //利润在40万元至60万时的奖金else if (i<=1000000)bonus=bon6+(i-600000)*0.015; //利润在60万元至100万时的奖金elsebonus=bon10+(i-1000000)*0.01; //利润在100万元以上时的奖金cout<<"bonus="<<bonus<<endl;return 0;}3.13题另一解#include <iostream>using namespace std;int main (){long i;float bonus,bon1,bon2,bon4,bon6,bon10; int c;bon1=100000*0.1;bon2=bon1+100000*0.075;bon4=bon2+200000*0.05;bon6=bon4+200000*0.03;bon10=bon6+400000*0.015;cout<<"enter i:";cin>>i;c=i/100000;if (c>10) c=10;switch(c){case 0: bonus=i*0.1; break;case 1: bonus=bon1+(i-100000)*0.075; break;case 2:case3:bonus=bon2+(i-200000)*0.05;break;case 4:case5:bonus=bon4+(i-400000)*0.03;break;case 6:case 7:case 8:case 9: bonus=bon6+(i-600000)*0.015; break;case 10: bonus=bon10+(i-1000000)*0.01;}cout<<"bonus="<<bonus<<endl;return 0;}3.14题#include <iostream>using namespace std;int main (){int t,a,b,c,d;cout<<"enter four numbers:";cin>>a>>b>>c>>d;cout<<"a="<<a<<", b="<<b<<", c="<<c<<",d="<<d<<endl;if (a>b){t=a;a=b;b=t;}if (a>c){t=a; a=c; c=t;}if (a>d){t=a; a=d; d=t;}if (b>c){t=b; b=c; c=t;}if (b>d){t=b; b=d; d=t;}if (c>d){t=c; c=d; d=t;}cout<<"the sorted sequence:"<<endl;cout<<a<<", "<<b<<", "<<c<<", "<<d<<endl; return 0;}3.15题#include <iostream>using namespace std;int main (){int p,r,n,m,temp;cout<<"please enter two positive integer numbers n,m:";cin>>n>>m;if (n<m){temp=n;n=m;m=temp; //把大数放在n中, 小数放在m中}p=n*m; //先将n和m的乘积保存在p中, 以便求最小公倍数时用while (m!=0) //求n和m的最大公约数{r=n%m;n=m;m=r;}cout<<"HCF="<<n<<endl;cout<<"LCD="<<p/n<<endl; // p是原来两个整数的乘积return 0;}3.16题#include <iostream>using namespace std;int main (){char c;int letters=0,space=0,digit=0,other=0;cout<<"enter one line::"<<endl;while((c=getchar())!='\n'){if (c>='a' && c<='z'||c>='A' && c<='Z')letters++;else if (c==' ')space++;else if (c>='0' && c<='9')digit++;elseother++;}cout<<"letter:"<<letters<<", space:"<<space<<", digit:"<<digit<<", other:"<<other<<endl;return 0;}3.17题#include <iostream>using namespace std;int main (){int a,n,i=1,sn=0,tn=0;cout<<"a,n=:";cin>>a>>n;while (i<=n){tn=tn+a; //赋值后的tn为i个a 组成数的值sn=sn+tn; //赋值后的sn为多项式前i项之和a=a*10;++i;}cout<<"a+aa+aaa+...="<<sn<<endl;return 0;}3.18题#include <iostream>using namespace std;int main (){float s=0,t=1;int n;for (n=1;n<=20;n++){t=t*n; // 求n!s=s+t; // 将各项累加}cout<<"1!+2!+...+20!="<<s<<endl;return 0;}3.19题#include <iostream>using namespace std;int main (){int i,j,k,n;cout<<"narcissus numbers are:"<<endl;for (n=100;n<1000;n++){i=n/100;j=n/10-i*10;k=n%10;if (n == i*i*i + j*j*j + k*k*k)cout<<n<<" ";}cout<<endl;return 0;}3.20题#include <iostream>using namespace std;int main(){const int m=1000; // 定义寻找范围int k1,k2,k3,k4,k5,k6,k7,k8,k9,k10;int i,a,n,s;for (a=2;a<=m;a++) // a是2~1000之间的整数,检查它是否为完数{n=0; // n用来累计a的因子的个数s=a; // s用来存放尚未求出的因子之和,开始时等于afor (i=1;i<a;i++) // 检查i是否为a 的因子if (a%i==0) // 如果i是a的因子{n++; // n加1,表示新找到一个因子s=s-i; // s减去已找到的因子,s的新值是尚未求出的因子之和switch(n) // 将找到的因子赋给k1,...,k10{case 1:k1=i; break; // 找出的笫1个因子赋给k1case 2:k2=i; break; // 找出的笫2个因子赋给k2case 3:k3=i; break; // 找出的笫3个因子赋给k3case 4:k4=i; break; // 找出的笫4个因子赋给k4case 5:k5=i; break; // 找出的笫5个因子赋给k5case 6:k6=i; break; // 找出的笫6个因子赋给k6case 7:k7=i; break; // 找出的笫7个因子赋给k7case 8:k8=i; break; // 找出的笫8个因子赋给k8case 9:k9=i; break; // 找出的笫9个因子赋给k9case 10:k10=i; break; // 找出的笫10个因子赋给k10}}if (s==0) // s=0表示全部因子都已找到了{cout<<a<<" is a 完数"<<endl;cout<<"its factors are:";if (n>1) cout<<k1<<","<<k2; // n>1表示a至少有2个因子if (n>2) cout<<","<<k3; // n>2表示至少有3个因子,故应再输出一个因子if (n>3) cout<<","<<k4; // n>3表示至少有4个因子,故应再输出一个因子if (n>4) cout<<","<<k5; // 以下类似if (n>5) cout<<","<<k6;if (n>6) cout<<","<<k7;if (n>7) cout<<","<<k8;if (n>8) cout<<","<<k9;if (n>9) cout<<","<<k10;cout<<endl<<endl;}}return 0;}3.20题另一解#include <iostream>using namespace std;int main(){int m,s,i;for (m=2;m<1000;m++){s=0;for (i=1;i<m;i++)if ((m%i)==0) s=s+i;if(s==m){cout<<m<<" is a完数"<<endl;cout<<"its factors are:";for (i=1;i<m;i++)if (m%i==0) cout<<i<<" ";cout<<endl;}}return 0;}3.20题另一解#include <iostream>using namespace std;int main(){int k[11];int i,a,n,s;for (a=2;a<=1000;a++){n=0;s=a;for (i=1;i<a;i++)if ((a%i)==0){n++;s=s-i;k[n]=i; // 将找到的因子赋给k[1]┅k[10]}if (s==0){cout<<a<<" is a 完数"<<endl;cout<<"its factors are:";for (i=1;i<n;i++)cout<<k[i]<<" ";cout<<k[n]<<endl;}}return 0;}3.21题#include <iostream>using namespace std;int main(){int i,t,n=20;double a=2,b=1,s=0;for (i=1;i<=n;i++){s=s+a/b;t=a;a=a+b; // 将前一项分子与分母之和作为下一项的分子b=t; // 将前一项的分子作为下一项的分母}cout<<"sum="<<s<<endl;return 0;} 3.22题#include <iostream>using namespace std;int main(){int day,x1,x2;day=9;x2=1;while(day>0){x1=(x2+1)*2; // 第1天的桃子数是第2天桃子数加1后的2倍x2=x1;day--;}cout<<"total="<<x1<<endl;return 0;}3.23题#include <iostream>#include <cmath>using namespace std;int main(){float a,x0,x1;cout<<"enter a positive number:"; cin>>a; // 输入a的值x0=a/2;x1=(x0+a/x0)/2;do{x0=x1;x1=(x0+a/x0)/2;}while(fabs(x0-x1)>=1e-5);cout<<"The square root of "<<a<<" is "<<x1<<endl;return 0;}3.24题#include <iostream>using namespace std;int main(){int i,k;for (i=0;i<=3;i++) // 输出上面4行*号{for (k=0;k<=2*i;k++)cout<<"*"; // 输出*号cout<<endl; //输出完一行*号后换行}for (i=0;i<=2;i++) // 输出下面3行*号{for (k=0;k<=4-2*i;k++)cout<<"*"; // 输出*号cout<<endl; // 输出完一行*号后换行}return 0;}3.25题#include <iostream>using namespace std;int main(){char i,j,k; /* i是a的对手;j是b的对手;k是c的对手*/for (i='X';i<='Z';i++)for (j='X';j<='Z';j++)if (i!=j)for (k='X';k<='Z';k++)if (i!=k && j!=k)if (i!='X' && k!='X' && k!='Z')cout<<"A--"<<i<<"B--"<<j<<" C--"<<k<<endl;return 0;}第四章4.1题#include <iostream>using namespace std;int main(){int hcf(int,int);int lcd(int,int,int);int u,v,h,l;cin>>u>>v;h=hcf(u,v);cout<<"H.C.F="<<h<<endl;l=lcd(u,v,h);cout<<"L.C.D="<<l<<endl;return 0;}int hcf(int u,int v){int t,r;if (v>u){t=u;u=v;v=t;}while ((r=u%v)!=0){u=v;v=r;}return(v);}int lcd(int u,int v,int h){return(u*v/h);}4.2题#include <iostream>#include <math.h>using namespace std;float x1,x2,disc,p,q;int main(){void greater_than_zero(float,float); void equal_to_zero(float,float);void smaller_than_zero(float,float); float a,b,c;cout<<"input a,b,c:";cin>>a>>b>>c;disc=b*b-4*a*c;cout<<"root:"<<endl;if (disc>0){greater_than_zero(a,b);cout<<"x1="<<x1<<",x2="<<x2<<endl; }else if (disc==0){equal_to_zero(a,b);cout<<"x1="<<x1<<",x2="<<x2<<endl;}else{smaller_than_zero(a,b);cout<<"x1="<<p<<"+"<<q<<"i"<<endl;cout<<"x2="<<p<<"-"<<q<<"i"<<endl;}return 0;}void greater_than_zero(float a,float b) /* 定义一个函数,用来求disc>0时方程的根*/{x1=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);}void equal_to_zero(float a,float b) /* 定义一个函数,用来求disc=0时方程的根*/{x1=x2=(-b)/(2*a);}void smaller_than_zero(float a,float b) /* 定义一个函数,用来求disc<0时方程的根*/{p=-b/(2*a);q=sqrt(-disc)/(2*a);}4.3题#include <iostream>using namespace std;int main(){int prime(int); /* 函数原型声明*/int n;cout<<"input an integer:";cin>>n;if (prime(n))cout<<n<<" is a prime."<<endl;elsecout<<n<<" is not a prime."<<endl;return 0;}int prime(int n){int flag=1,i;for (i=2;i<n/2 && flag==1;i++)if (n%i==0)flag=0;return(flag);}4.4题#include <iostream>using namespace std;int main(){int fac(int);int a,b,c,sum=0;cout<<"enter a,b,c:";cin>>a>>b>>c;sum=sum+fac(a)+fac(b)+fac(c);cout<<a<<"!+"<<b<<"!+"<<c<<"!="<<sum<<e ndl;return 0;}int fac(int n){int f=1;for (int i=1;i<=n;i++)f=f*i;return f;}4.5题#include <iostream>#include <cmath>using namespace std;int main(){double e(double);double x,sinh;cout<<"enter x:";cin>>x;sinh=(e(x)+e(-x))/2;cout<<"sinh("<<x<<")="<<sinh<<endl;return 0;}double e(double x){return exp(x);}4.6题#include <iostream>#include <cmath>using namespace std;int main(){doublesolut(double ,double ,double ,double ); double a,b,c,d;cout<<"input a,b,c,d:";cin>>a>>b>>c>>d;cout<<"x="<<solut(a,b,c,d)<<endl;return 0;}double solut(double a,double b,double c,double d){double x=1,x0,f,f1;do{x0=x;f=((a*x0+b)*x0+c)*x0+d;f1=(3*a*x0+2*b)*x0+c;x=x0-f/f1;}while(fabs(x-x0)>=1e-5);return(x);}4.7题#include <iostream>#include <cmath>using namespace std;int main(){void godbaha(int);int n;cout<<"input n:";cin>>n;godbaha(n);return 0;}void godbaha(int n){int prime(int);int a,b;for(a=3;a<=n/2;a=a+2){if(prime(a)){b=n-a;if (prime(b))cout<<n<<"="<<a<<"+"<<b<<endl;}}}int prime(int m){int i,k=sqrt(m);for(i=2;i<=k;i++)if(m%i==0) break;if (i>k) return 1;else return 0;}4.8题#include <iostream>using namespace std;int main(){int x,n;float p(int,int);cout<<"input n & x:";cin>>n>>x;cout<<"n="<<n<<",x="<<x<<endl;;cout<<"P"<<n<<"(x)="<<p(n,x)<<endl; return 0;}float p(int n,int x){if (n==0)return(1);else if (n==1)return(x);elsereturn(((2*n-1)*x*p((n-1),x)-(n-1)*p((n-2),x))/n);}4.9题#include <iostream>using namespace std;int main(){void hanoi(int n,char one,char two,char three);int m;cout<<"input the number of diskes:"; cin>>m;cout<<"The steps of moving "<<m<<" disks:"<<endl;hanoi(m,'A','B','C');return 0;}void hanoi(int n,char one,char two,char three) //将n个盘从one座借助two座,移到three座{void move(char x,char y);if(n==1) move(one,three);else{hanoi(n-1,one,three,two);move(one,three);hanoi(n-1,two,one,three);}}void move(char x,char y){cout<<x<<"-->"<<y<<endl;}4.10题#include <iostream>using namespace std;int main(){void convert(int n);int number;cout<<"input an integer:";cin>>number;cout<<"output:"<<endl;if (number<0){cout<<"-";number=-number;}convert(number);cout<<endl;return 0;}void convert(int n){int i;char c;if ((i=n/10)!=0)convert(i);c=n%10+'0';cout<<" "<<c;}4.11题#include <iostream>using namespace std;int main(){int f(int);int n,s;cout<<"input the number n:";cin>>n;s=f(n);cout<<"The result is "<<s<<endl;return 0;}int f(int n){;if (n==1)return 1;elsereturn (n*n+f(n-1));}4.12题#include <iostream>#include <cmath>using namespace std;#define S(a,b,c) (a+b+c)/2#define AREA(a,b,c)sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(S(a,b,c) -c))int main(){float a,b,c;cout<<"input a,b,c:";cin>>a>>b>>c;if (a+b>c && a+c>b && b+c>a)cout<<"area="<<AREA(a,b,c)<<endl; elsecout<<"It is not a triangle!"<<endl; return 0;}4.14题#include <iostream>using namespace std;//#define LETTER 1int main(){char c;cin>>c;#if LETTERif(c>='a' && c<='z')c=c-32;#elseif(c>='A' && c<='Z')c=c+32;#endifcout<<c<<endl;return 0;}4.15题#include <iostream>using namespace std;#define CHANGE 1int main(){char ch[40];cout<<"input text:"<<endl;;gets(ch);#if (CHANGE){for (int i=0;i<40;i++){if (ch[i]!='\0')if (ch[i]>='a'&& ch[i]<'z'||ch[i]>'A'&& ch[i]<'Z')ch[i]+=1;else if (ch[i]=='z'||ch[i]=='Z')ch[i]-=25;}}#endifcout<<"output:"<<endl<<ch<<endl;return 0;}4.16题file#include <iostream>using namespace std;int a;int main(){extern int power(int);int b=3,c,d,m;cout<<"enter an integer a and its power m:"<<endl;cin>>a>>m;c=a*b;cout<<a<<"*"<<b<<"="<<c<<endl;d=power(m);cout<<a<<"**"<<m<<"="<<d<<endl; return 0;}4.16题fileextern int a;int power(int n){int i,y=1;for(i=1;i<=n;i++)y*=a;return y;}第五章5.1题#include <iostream>#include <iomanip>using namespace std;#include <math.h>int main(){int i,j,n,a[101];for (i=1;i<=100;i++)a[i]=i;a[1]=0;for (i=2;i<sqrt(100);i++)for (j=i+1;j<=100;j++){if(a[i]!=0 && a[j]!=0)if (a[j]%a[i]==0)a[j]=0; }cout<<endl;for (i=1,n=0;i<=100;i++){if (a[i]!=0){cout<<setw(5)<<a[i]<<" ";n++;}if(n==10){cout<<endl;n=0;}}cout<<endl;return 0;}5.2题#include <iostream>using namespace std;//#include <math.h>int main(){int i,j,min,temp,a[11];cout<<"enter data:"<<endl;for (i=1;i<=10;i++){cout<<"a["<<i<<"]=";cin>>a[i]; //输入10个数}cout<<endl<<"The original numbers:"<<endl;;for (i=1;i<=10;i++)cout<<a[i]<<" "; // 输出这10个数cout<<endl;;for (i=1;i<=9;i++) //以下8行是对10个数排序{min=i;for (j=i+1;j<=10;j++)if (a[min]>a[j]) min=j;temp=a[i]; //以下3行将a[i+1]~a[10]中最小者与a[i] 对换a[i]=a[min];a[min]=temp;}cout<<endl<<"The sorted numbers:"<<endl;for (i=1;i<=10;i++) // 输出已排好序的10个数cout<<a[i]<<" ";cout<<endl;return 0;}5.3题#include <iostream>using namespace std;int main(){int a[3][3],sum=0;int i,j;cout<<"enter data:"<<endl;;for (i=0;i<3;i++)for (j=0;j<3;j++)cin>>a[i][j];for (i=0;i<3;i++)sum=sum+a[i][i];cout<<"sum="<<sum<<endl;return 0;}5.4题#include <iostream>using namespace std;int main(){int a[11]={1,4,6,9,13,16,19,28,40,100};int num,i,j;cout<<"array a:"<<endl;for (i=0;i<10;i++)cout<<a[i]<<" ";cout<<endl;;cout<<"insert data:";cin>>num;if (num>a[9])a[10]=num;else。

c 程序设计谭浩强课后习题答案

c  程序设计谭浩强课后习题答案

c 程序设计谭浩强课后习题答案C程序设计谭浩强课后习题答案谭浩强的《C程序设计》是一本经典的C语言教材,它系统地介绍了C语言的基本语法和常用编程技巧。

在学习过程中,课后习题是巩固知识和提高编程能力的重要环节。

本文将为大家提供一些C程序设计谭浩强课后习题的答案,希望对大家的学习有所帮助。

1. 编写一个程序,输入一个整数,判断该整数是奇数还是偶数。

```c#include <stdio.h>int main(){int num;printf("请输入一个整数:");scanf("%d", &num);if (num % 2 == 0){printf("%d是偶数。

\n", num);}else{printf("%d是奇数。

\n", num);}return 0;}```2. 编写一个程序,输入一个字符,判断该字符是大写字母、小写字母还是其他字符。

```c#include <stdio.h>int main(){char ch;printf("请输入一个字符:");scanf("%c", &ch);if (ch >= 'A' && ch <= 'Z'){printf("%c是大写字母。

\n", ch);}else if (ch >= 'a' && ch <= 'z')printf("%c是小写字母。

\n", ch);}else{printf("%c是其他字符。

\n", ch);}return 0;}```3. 编写一个程序,输入一个年份,判断该年份是否是闰年。

```c#include <stdio.h>int main(){int year;printf("请输入一个年份:");scanf("%d", &year);if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)printf("%d年是闰年。

c程序设计谭浩强习题答案

c程序设计谭浩强习题答案

c程序设计谭浩强习题答案C程序设计是计算机科学与技术专业学生必修的一门课程,谭浩强教授编写的《C程序设计》教材广泛用于教学中,其习题集也是学生学习过程中不可或缺的练习材料。

以下是针对该教材习题的一些参考答案,供参考使用。

第一章:C语言概述1. 问题1:解释C语言的特点。

- 答案:C语言是一种通用的、过程式的编程语言,具有以下特点:简洁、高效、可移植性高、结构化、支持多种编程范式等。

2. 问题2:C语言的发展历史。

- 答案:C语言由丹尼斯·里奇在1972年开发,最初用于UNIX操作系统的编写。

随着UNIX系统的流行,C语言也逐渐被广泛使用并发展。

第二章:数据类型、运算符和表达式1. 问题1:C语言中基本数据类型有哪些?- 答案:C语言中的基本数据类型包括整型(int)、字符型(char)、浮点型(float和double)等。

2. 问题3:解释运算符的优先级。

- 答案:运算符优先级决定了表达式中各运算符的计算顺序。

例如,乘除运算符优先于加减运算符。

第三章:控制语句1. 问题1:解释if语句的用法。

- 答案:if语句用于条件判断,根据条件是否为真来决定执行不同的代码块。

2. 问题2:while循环和for循环的区别是什么?- 答案:while循环在每次迭代前检查条件,而for循环在循环开始前初始化,然后在每次迭代后检查条件。

第四章:数组1. 问题1:一维数组和多维数组的区别。

- 答案:一维数组是线性的,只有一个索引;多维数组可以看作是数组的数组,具有多个索引。

2. 问题2:数组的初始化方法。

- 答案:数组可以通过赋值语句逐个初始化,或者使用初始化列表在声明时初始化。

第五章:指针1. 问题1:指针是什么?- 答案:指针是一种变量,它存储了另一个变量的内存地址。

2. 问题2:指针和数组的关系。

- 答案:数组名可以作为指向数组首元素的指针使用,指针可以用于遍历数组。

第六章:函数1. 问题1:函数的定义和声明。

C语言程序设计第三版谭浩强课后习题答案完整版

C语言程序设计第三版谭浩强课后习题答案完整版

C语言程序设计第三版谭浩强课后习题答案完整版第一章1.5请参照本章例题,编写一个C程序,输出以下信息:**************************Very Good!**************************解:mian(){printf(“**************************”);printf(“\n”);printf(“Very Good!\n”);printf(“\n”);printf(“**************************”);}1.6 编写一个程序,输入a、b、c三个值,输出其中最大值。

解:mian(){int a,b,c,max;printf(“请输入三个数a,b,c:\n”);scanf(“%d,%d,%d”,&a,&b,&c);max=a;if(max<b)max=b;if(max<c)max=c;printf(“最大数为:“%d”,max);}第三章3.6写出以下程序运行的结果。

main(){char c1=’a’,c2=’b’,c3=’c’,c4=’\101’,c5=’\116’;printf(“a%cb%c\tc%c\tabc\n”,c1,c2,c3);printf(“\t\b%c %c”,c4,c5);}解:aaㄩbbㄩㄩㄩccㄩㄩㄩㄩㄩㄩabcAㄩN3.7要将"China"译成密码,译码规律是:用原来字母后面的第4个字母代替原来的字母.例如,字母"A"后面第4个字母是"E"."E"代替"A"。

因此,"China"应译为"Glmre"。

请编一程序,用赋初值的方法使cl、c2、c3、c4、c5五个变量的值分别为,’C’、’h’、’i’、’n’、’a’,经过运算,使c1、c2、c3、c4、c5分别变为’G’、’l’、’m’、’r’、’e’,并输出。

C程序设计(谭浩强第三版)课后答案完整版

C程序设计(谭浩强第三版)课后答案完整版
{ )(niam )IICSA(89_ro_ˊbˊ=2c )IICSA(79_ro_ˊaˊ=1c 04.2-=x+z_ _02.1-=z+y_ _006.3_=y+x 000006.3-=z,000004.2=y,000002.1=x 5_=c_ _4_=b_ _3_=a 。 �出输计设和型类量变义定括包�序程出写请�果结和式格出输下以到得想 。 ’b’=2c� ’a’=1c�567821=n�47215=u�6.3-=z�4.2=y�2.1=x�5=c�4=b�3=a 若 4�4 章四第 0 )6( 0 )5( 0 )4( 06 )3( 01 )2( 42 )1( �解 a=*a=-a=+a�6� 5 于等值的 n�)2=%n(=%a �5� a+a=/a�4� 3+2=*a �3� 2=-a �2� a=+a�1� 。量变型整为义定已都 n 和 a 设。21=a 来原设�值的 a 后算运式达表面下出写 21.3 01�9�11�9 �解 } ;)n,m,j,i,”d%,d%,d%,d%“(ftnirp ;++j=n ;i++=m ;01=j ;8=i ;n,m,j,i tni{ )(niam 。果结的行运序程出写 01.3 5.3�2� 5.2�1� 5.2=y,5.3=x,3=b,2=a 设 y)tni(%x)tni(+2/)b+a()taolf(�2� 7.4=y,7=a,5.2=x 设 4/2%)y+x()tni(*3%a+x�1� 。值的式达表术算面下求 9.3 ermlG 是码密 �果结行运 } ;)5c,4c,3c,2c,1c,"n\c%c%c%c%c%是码密"(ftnirp ;4=+5c ;4=+4c ;4=+3c ;4=+2c ;4=+1c ;’a’=5c,’n’=4c,’i’=3c,’h’=2c,’C’=1c rahc {

谭浩强C程序设计第三版课后答案(整本书).doc

谭浩强C程序设计第三版课后答案(整本书).doc
scanf(“a=%d b=%d”,&a , &b);
scanf(“%f %e”,&x , &y);
scanf(“%c %c”,&c1 , &c2);
}
解:
a=3□b=7
□8.5□71.82
□A□a
4.7
scanf(“%5d%5d%c%c%f%f%*f,%f”,&a,&b,&c1,&c2,&x,&y,&z);
(7)(-28654)10=(110022)8=(9012)16
(8)(21003)10=(51013)8=(520B)16
3
变量的类型
25
-2
32769
int型
long型
sort型
signed char(8位)
unsigned int型
unsigned long型
unsigned short型
unsigned char型
unsigned u=65535;
printf(“%d%d\n”,a,b);
printf(“%3d%3d\n”,a,b);
printf(“%f,%f\n”,x,y);
printf(“%-10f,%-10f\n”,x,y);
printf(“%8.2f, %8.2f,%4f, %4f, %3f, %3f\n”,x,y,x,y,x,y);
voidmain ( )
{ char c1=’a’, c2=’b’, c3=’c’, c4=’\101’, c5=’\116’;
printf (“a%c b%c\t c%c\t abc\n”, c1, c2, c3);
printf (“\t\b%c %c”, c4, c5);

谭浩强C程序设计第三版课后答案(整本书).doc

谭浩强C程序设计第三版课后答案(整本书).doc
unsigned u=65535;
printf(“%d%d\n”,a,b);
printf(“%3d%3d\n”,a,b);
printf(“%f,%f\n”,x,y);
printf(“%-10f,%-10f\n”,x,y);
printf(“%8.2f, %8.2f,%4f, %4f, %3f, %3f\n”,x,y,x,y,x,y);
}
运行情况如下:
□□□10□□□20Aa1.5–3.75□1.5,67.8(此行为输入的数据,其中□为空格)
a=10, b=20, c1=A, c2=a, x=1.50, y=-3.75, z=67.80(此行为输出)
说明:按%5d格式的要求输入a和b时,要先键入三个空格,然后再键入10与20。%*f是用来禁止赋值的。在输入时,对应于%*f的地方,随意打入了一个数1.5,该值不会赋给任何变量。
此题可以这样考虑:带符号数在计算机中采用补码表示,正数的补码与原码相同,负数的补码=模+真值。若使用16位存储,模为216=65536。-617的补码为65536+(-167)=64919=(176627)8=(FD97)16
(5)(-111)10=(177621)8=(FF91)16
(6)(2483)10=(4663)8=(9B3)16
}
运行结果:
5□7
□□5□□7
67.856400, -789.124023
67.856400□, -789.124023
□□□67.86,□□-789.12,67.8564,-789.1240,67.856400,-789.124023
6.785640e+01,□□-7.9e+02
A,65,101,41

C程序设计谭浩强_课后习题答案_共11章

C程序设计谭浩强_课后习题答案_共11章

1-5 #include <>int main ( ){ printf ("**************************\n\n"); printf(" Very Good!\n\n");printf ("**************************\n"); return 0;}1-6#include <>int main(){int a,b,c,max;printf("please input a,b,c:\n");scanf("%d,%d,%d",&a,&b,&c);max=a;if (max<b)max=b;if (max<c)max=c;printf("The largest number is %d\n",max);return 0;}3-1 #include <>#include <>int main(){float p,r,n;r=;n=10;p=pow(1+r,n);printf("p=%f\n",p);return 0;}3-2-1#include <>#include <>int main(){float r5,r3,r2,r1,r0,p,p1,p2,p3,p4,p5;p=1000;r5=;r3=;r2=;r1=;r0=;p1=p*((1+r5)*5); #include <>#include <>int main(){float d=300000,p=6000,r=,m;m=log10(p/(p-d*r))/log10(1+r);printf("m=%\n",m);return 0;}3-4#include <>int main(){int c1,c2;c1=197;c2=198;printf("c1=%c,c2=%c\n",c1,c2);printf("c1=%d,c2=%d\n",c1,c2);return 0;}3-5#include <>int main(){int a,b;float x,y;char c1,c2;scanf("a=%d b=%d",&a,&b);scanf("%f %e",&x,&y);scanf("%c%c",&c1,&c2);printf("a=%d,b=%d,x=%f,y=%f,c1=%c,c2=%c\n",a,b,x,y,c1,c2); return 0;}3-6#include <>int main(){char c1='C',c2='h',c3='i',c4='n',c5='a';c1=c1+4;c2=c2+4;c3=c3+4;c4=c4+4;c5=c5+4;printf("passwor is %c%c%c%c%c\n",c1,c2,c3,c4,c5);return 0;}3-7#include <>int main (){float h,r,l,s,sq,vq,vz;float pi=;printf("请输入圆半径r,圆柱高h∶");scanf("%f,%f",&r,&h); #include <> int main(){ int x,y;printf("输入x:");scanf("%d",&x);if(x<1) /* x<1 */{ y=x;printf("x=%3d, y=x=%d\n" ,x,y);}else if(x<10) /* 1=<x<10 */{ y=2*x-1;printf("x=%d, y=2*x-1=%d\n",x,y);}else /* x>=10 */{ y=3*x-11;printf("x=%d, y=3*x-11=%d\n",x,y);}return 0;}4-8#include <>int main(){ float score;char grade;printf("请输入学生成绩:");scanf("%f",&score);while (score>100||score<0){printf("\n 输入有误,请重输");scanf("%f",&score);}switch((int)(score/10)){case 10:case 9: grade='A';break;case 8: grade='B';break;case 7: grade='C';break;case 6: grade='D';break;case 5:case 4:case 3:case 2:case 1:case 0: grade='E';}printf("成绩是 %,相应的等级是%c\n ",score,grade);return 0;}4-10-1#include <>int main(){int i;double bonus,bon1,bon2,bon4,bon6,bon10;bon1=100000*;bon2=bon1+100000*;bon4=bon2+100000*;bon6=bon4+100000*;bon10=bon6+400000*;printf("请输入利润i:");scanf("%d",&i);if (i<=100000)bonus=i*;else if (i<=200000)bonus=bon1+(i-100000)*;else if (i<=400000)bonus=bon2+(i-200000)*;else if (i<=600000)bonus=bon4+(i-400000)*;else if (i<=1000000)bonus=bon6+(i-600000)*;elsebonus=bon10+(i-1000000)*;printf("奖金是: %\n",bonus);return 0;}4-10-2#include <>int main(){int i;double bonus,bon1,bon2,bon4,bon6,bon10;int branch;bon1=100000*;bon2=bon1+100000*;bon4=bon2+200000*;bon6=bon4+200000*;bon10=bon6+400000*;printf("请输入利润i:");scanf("%d",&i);branch=i/100000;if (branch>10) branch=10;switch(branch){ case 0:bonus=i*;break;case 1:bonus=bon1+(i-100000)*;break; case 2:case 3: bonus=bon2+(i-200000)*;break; case 4:case 5: bonus=bon4+(i-400000)*;break; case 6:case 7:case 8:case 9: bonus=bon6+(i-600000)*;break; case 10: bonus=bon10+(i-1000000)*;}printf("奖金是 %\n",bonus);return 0;}4-11#include <>int main(){int t,a,b,c,d;printf("请输入四个数:");scanf("%d,%d,%d,%d",&a,&b,&c,&d);printf("a=%d,b=%d,c=%d,d=%d\n",a,b,c,d); if (a>b){ t=a;a=b;b=t;}if (a>c){ t=a;a=c;c=t;}if (a>d){ t=a;a=d;d=t;}if (b>c){ t=b;b=c;c=t;}if (b>d){ t=b;b=d;d=t;}if (c>d){ t=c;c=d;d=t;}printf("排序结果如下: \n");printf("%d %d %d %d \n" ,a,b,c,d); return 0;}4-12#include <>int main(){int h=10;float x1=2,y1=2,x2=-2,y2=2,x3=-2,y3=-2,x4=2,y4=-2,x,y,d1,d2,d3,d4; printf("请输入一个点(x,y):");scanf("%f,%f",&x,&y);d1=(x-x4)*(x-x4)+(y-y4)*(y-y4); /*求该点到各中心点距离*/ d2=(x-x1)*(x-x1)+(y-y1)*(y-y1);d3=(x-x2)*(x-x2)+(y-y2)*(y-y2);d4=(x-x3)*(x-x3)+(y-y3)*(y-y3);if (d1>1 && d2>1 && d3>1 && d4>1) h=0; /*判断该点是否在塔外*/ printf("该点高度为 %d\n",h);return 0;}5-3#include <>int main(){int p,r,n,m,temp;printf("请输入两个正整数n,m:");scanf("%d,%d,",&n,&m);if (n<m){temp=n;n=m;m=temp;}p=n*m;while(m!=0){r=n%m;n=m;m=r;}printf("它们的最大公约数为:%d\n",n);printf("它们的最小公约数为:%d\n",p/n);return 0;}5-4#include <>int main(){char c;int letters=0,space=0,digit=0,other=0;printf("请输入一行字符:\n");while((c=getchar())!='\n'){if (c>='a' && c<='z' || c>='A' && c<='Z')letters++;else if (c==' ')space++;else if (c>='0' && c<='9')digit++;elseother++;}printf("字母数:%d\n空格数:%d\n数字数:%d\n其它字符数:%d\n",letters,space,digit,other);return 0;}5-5#include <>int main(){int a,n,i=1,sn=0,tn=0;printf("a,n=:");scanf("%d,%d",&a,&n);while (i<=n){tn=tn+a; /*赋值后的tn为i个 a组成数的值*/sn=sn+tn; /*赋值后的sn为多项式前i项之和*/a=a*10;++i;}printf("a+aa+aaa+...=%d\n",sn);return 0;}5-6#include <>int main(){double s=0,t=1;int n;for (n=1;n<=20;n++){t=t*n;s=s+t;}printf("1!+2!+...+20!=%\n",s);return 0;}5-7#include <>int main(){int n1=100,n2=50,n3=10;double k,s1=0,s2=0,s3=0;for (k=1;k<=n1;k++) /*计算1到100的和*/{s1=s1+k;}for (k=1;k<=n2;k++) /*计算1到50各数的平方和*/ {s2=s2+k*k;}for (k=1;k<=n3;k++) /*计算1到10的各倒数和*/ {s3=s3+1/k;}printf("sum=%\n",s1+s2+s3);return 0;}5-8#include <>int main(){int i,j,k,n;printf("parcissus numbers are ");for (n=100;n<1000;n++){i=n/100;j=n/10-i*10;k=n%10;if (n==i*i*i + j*j*j + k*k*k)printf("%d ",n);}printf("\n");return 0;}5-9#include <>int main(){int m,s,i;for (m=2;m<1000;m++){s=0;for (i=1;i<m;i++)if ((m%i)==0) s=s+i;if(s==m){printf("%d,its factors are ",m);for (i=1;i<m;i++)if (m%i==0) printf("%d ",i);printf("\n");}}return 0;}5-10#include <>int main(){int i,n=20;double a=2,b=1,s=0,t;for (i=1;i<=n;i++){s=s+a/b;t=a,a=a+b,b=t;}printf("sum=%\n",s);return 0;}5-11#include <>int main(){double sn=100,hn=sn/2;int n;for (n=2;n<=10;n++){sn=sn+2*hn; /*第n次落地时共经过的米数*/ hn=hn/2; /*第n次反跳高度*/}printf("第10次落地时共经过%f米\n",sn);printf("第10次反弹%f米\n",hn);return 0;}5-12#include <>int main(){int day,x1,x2;day=9;x2=1;while(day>0){x1=(x2+1)*2; /*第1天的桃子数是第2天桃子数加1后的2倍.*/ x2=x1;day--;}printf("total=%d\n",x1);return 0;}5-13#include <>#include <>int main(){float a,x0,x1;printf("enter a positive number:");scanf("%f",&a);x0=a/2;x1=(x0+a/x0)/2;do{x0=x1;x1=(x0+a/x0)/2;}while(fabs(x0-x1)>=1e-5);printf("The square root of % is %\n",a,x1);return 0;}5-14#include <>#include <>int main(){double x1,x0,f,f1;x1=;do{x0=x1;f=((2*x0-4)*x0+3)*x0-6;f1=(6*x0-8)*x0+3;x1=x0-f/f1;}while(fabs(x1-x0)>=1e-5);printf("The root of equation is %\n",x1);return 0;}5-15#include <>#include <>int main(){float x0,x1,x2,fx0,fx1,fx2; do{printf("enter x1 & x2:"); scanf("%f,%f",&x1,&x2); fx1=x1*((2*x1-4)*x1+3)-6; fx2=x2*((2*x2-4)*x2+3)-6; }while(fx1*fx2>0);do{x0=(x1+x2)/2;fx0=x0*((2*x0-4)*x0+3)-6; if ((fx0*fx1)<0){x2=x0;fx2=fx0;}else{x1=x0;fx1=fx0;}}while(fabs (fx0)>=1e-5); printf("x=%\n",x0);return 0;}5-16#include <>int main(){int i,j,k;for (i=0;i<=3;i++){for (j=0;j<=2-i;j++)printf(" ");for (k=0;k<=2*i;k++)printf("*");printf("\n");}for (i=0;i<=2;i++){for (j=0;j<=i;j++)printf(" ");for (k=0;k<=4-2*i;k++)printf("*");printf("\n");}return 0;}5-17#include <>int main(){char i,j,k; /*是a的对手;j是b的对手;k是c的对手*/ for (i='x';i<='z';i++)for (j='x';j<='z';j++)if (i!=j)for (k='x';k<='z';k++)if (i!=k && j!=k)if (i!='x' && k!='x' && k!='z')printf("A--%c\nB--%c\nC--%c\n",i,j,k);return 0;}6-1#include <>#include <>int main(){int i,j,n,a[101];for (i=1;i<=100;i++)a[i]=i;a[1]=0;for (i=2;i<sqrt(100);i++)for (j=i+1;j<=100;j++){if(a[i]!=0 && a[j]!=0)if (a[j]%a[i]==0)a[j]=0;}printf("\n");for (i=2,n=0;i<=100;i++){ if(a[i]!=0){printf("%5d",a[i]);n++;}if(n==10){printf("\n");n=0;}}printf("\n");return 0;}6-2#include <>int main(){int i,j,min,temp,a[11];printf("enter data:\n");for (i=1;i<=10;i++){printf("a[%d]=",i);scanf("%d",&a[i]);}printf("\n");printf("The orginal numbers:\n"); for (i=1;i<=10;i++)printf("%5d",a[i]);printf("\n");for (i=1;i<=9;i++){min=i;for (j=i+1;j<=10;j++)if (a[min]>a[j]) min=j;temp=a[i];a[i]=a[min];a[min]=temp;}printf("\nThe sorted numbers:\n"); for (i=1;i<=10;i++)printf("%5d",a[i]);printf("\n");return 0;}6-3#include <>int main(){int a[3][3],sum=0;int i,j;printf("enter data:\n");for (i=0;i<3;i++)for (j=0;j<3;j++)scanf("%3d",&a[i][j]);for (i=0;i<3;i++)sum=sum+a[i][i];printf("sum=%6d\n",sum);return 0;}6-4#include <>int main(){ int a[11]={1,4,6,9,13,16,19,28,40,100};int temp1,temp2,number,end,i,j;printf("array a:\n");for (i=0;i<10;i++)printf("%5d",a[i]);printf("\n");printf("insert data:");scanf("%d",&number);end=a[9];if (number>end)a[10]=number;else{for (i=0;i<10;i++){if (a[i]>number){temp1=a[i];a[i]=number;for (j=i+1;j<11;j++){temp2=a[j];a[j]=temp1;temp1=temp2;}break;}}}printf("Now array a:\n");for (i=0;i<11;i++)printf("%5d",a[i]);printf("\n");return 0;}6-5#include <>#define N 5int main(){ int a[N],i,temp;printf("enter array a:\n");for (i=0;i<N;i++)scanf("%d",&a[i]);printf("array a:\n");for (i=0;i<N;i++)printf("%4d",a[i]);for (i=0;i<N/2;i++) n",number);;printf("continu or not(Y/N)");scanf(" %c",&c);if (c=='N'||c=='n')flag=0;}return 0;}6-10#include <>int main(){int i,j,upp,low,dig,spa,oth;char text[3][80];upp=low=dig=spa=oth=0;for (i=0;i<3;i++){ printf("please input line %d:\n",i+1);gets(text[i]);for (j=0;j<80 && text[i][j]!='\0';j++){if (text[i][j]>='A'&& text[i][j]<='Z') upp++;else if (text[i][j]>='a' && text[i][j]<='z') low++;else if (text[i][j]>='0' && text[i][j]<='9') dig++;else if (text[i][j]==' ')spa++;elseoth++;}}printf("\nupper case: %d\n",upp);printf("lower case: %d\n",low);printf("digit : %d\n",dig);printf("space : %d\n",spa);printf("other : %d\n",oth);return 0;}6-11#include <>int main(){ char a[5]={'*','*','*','*','*'};int i,j,k;char space=' ';for (i=0;i<5;i++){ printf("\n");printf(" ");for (j=1;j<=i;j++)printf("%c",space);for (k=0;k<5;k++)printf("%c",a[k]);}printf("\n");return 0;}6-12a-c#include <>int main(){ int j,n;char ch[80],tran[80];printf("input cipher code:");gets(ch);printf("\ncipher code :%s",ch);j=0;while (ch[j]!='\0'){ if ((ch[j]>='A') && (ch[j]<='Z'))tran[j]=155-ch[j];else if ((ch[j]>='a') && (ch[j]<='z')) tran[j]=219-ch[j];elsetran[j]=ch[j];j++;}n=j;printf("\noriginal text:");for (j=0;j<n;j++)putchar(tran[j]);printf("\n");return 0;}6-12b#include <>int main(){int j,n;char ch[80];printf("input cipher code:\n");gets(ch);printf("\ncipher code:%s\n",ch);j=0;while (ch[j]!='\0'){ if ((ch[j]>='A') && (ch[j]<='Z'))ch[j]=155-ch[j];else if ((ch[j]>='a') && (ch[j]<='z'))ch[j]=219-ch[j];elsech[j]=ch[j];j++;}n=j;printf("original text:");for (j=0;j<n;j++)putchar(ch[j]);printf("\n");return 0;}6-13#include <>int main(){ char s1[80],s2[40];int i=0,j=0;printf("input string1:");scanf("%s",s1);printf("input string2:");scanf("%s",s2);while (s1[i]!='\0')i++;while(s2[j]!='\0')s1[i++]=s2[j++];s1[i]='\0';printf("\nThe new string is:%s\n",s1);return 0;}6-14#include <>int main(){ int i,resu;char s1[100],s2[100];printf("input string1:");gets(s1);printf("\ninput string2:");gets(s2);i=0;while ((s1[i]==s2[i]) && (s1[i]!='\0'))i++; if (s1[i]=='\0' && s2[i]=='\0')resu=0;elseresu=s1[i]-s2[i];printf("\nresult:%d.\n",resu); return 0;}6-15#include <>#include <>int main(){ char s1[80],s2[80];int i;printf("input s2:");scanf("%s",s2);for (i=0;i<=strlen(s2);i++)s1[i]=s2[i];printf("s1:%s\n",s1);return 0;}7-1-1#include <>int main(){int hcf(int,int);int lcd(int,int,int);int u,v,h,l;scanf("%d,%d",&u,&v);h=hcf(u,v);printf("",h);l=lcd(u,v,h);printf("",l);return 0;}int hcf(int u,int v){int t,r;if (v>u){t=u;u=v;v=t;}while ((r=u%v)!=0){u=v;v=r;}return(v);}int lcd(int u,int v,int h){return(u*v/h);}7-1-2#include <>int Hcf,Lcd;int main(){void hcf(int,int);void lcd(int,int);int u,v;scanf("%d,%d",&u,&v);hcf(u,v);lcd(u,v);printf("",Hcf);printf("",Lcd);return 0;}void hcf(int u,int v){int t,r;if (v>u){t=u;u=v;v=t;}while ((r=u%v)!=0){u=v;v=r;}Hcf=v;}void lcd(int u,int v){Lcd=u*v/Hcf;}7-2#include <>#include <>float x1,x2,disc,p,q;int main(){void greater_than_zero(float,float);void equal_to_zero(float,float);void smaller_than_zero(float,float);float a,b,c;printf("input a,b,c:");scanf("%f,%f,%f",&a,&b,&c);printf("equation: %*x*x+%*x+%=0\n",a,b,c);disc=b*b-4*a*c;printf("root:\n");if (disc>0){greater_than_zero(a,b);printf("x1=%f\t\tx2=%f\n",x1,x2);}else if (disc==0){equal_to_zero(a,b);printf("x1=%f\t\tx2=%f\n",x1,x2);}else{smaller_than_zero(a,b);printf("x1=%f+%fi\tx2=%f-%fi\n",p,q,p,q); }return 0;}void greater_than_zero(float a,float b){x1=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);}void equal_to_zero(float a,float b){x1=x2=(-b)/(2*a);}void smaller_than_zero(float a,float b){p=-b/(2*a);q=sqrt(-disc)/(2*a);}7-3#include <>int main(){int prime(int);int n;printf("input an integer:");scanf("%d",&n);if (prime(n))printf("%d is a prime.\n",n);elseprintf("%d is not a prime.\n",n);return 0;}int prime(int n){int flag=1,i;for (i=2;i<n/2 && flag==1;i++) if (n%i==0)flag=0;return(flag);}7-4#include <>#define N 3int array[N][N];int main(){ void convert(int array[][3]); int i,j;printf("input array:\n");for (i=0;i<N;i++)for (j=0;j<N;j++)scanf("%d",&array[i][j]);printf("\noriginal array :\n"); for (i=0;i<N;i++){for (j=0;j<N;j++)printf("%5d",array[i][j]);printf("\n");}convert(array);printf("convert array:\n");for (i=0;i<N;i++){for (j=0;j<N;j++)printf("%5d",array[i][j]); printf("\n");}return 0;}void convert(int array[][3]){int i,j,t;for (i=0;i<N;i++)for (j=i+1;j<N;j++){t=array[i][j];array[i][j]=array[j][i];array[j][i]=t;}}#include <>#include <>int main(){void inverse(char str[]);char str[100];printf("input string:");scanf("%s",str);inverse(str);printf("inverse string:%s\n",str);return 0;}void inverse(char str[]){char t;int i,j;for (i=0,j=strlen(str);i<(strlen(str)/2);i++,j--){t=str[i];str[i]=str[j-1];str[j-1]=t;}}7-6#include <>int main(){void concatenate(char string1[],char string2[],char string[]); char s1[100],s2[100],s[100];printf("input string1:");scanf("%s",s1);printf("input string2:");scanf("%s",s2);concatenate(s1,s2,s);printf("\nThe new string is %s\n",s);return 0;}void concatenate(char string1[],char string2[],char string[]) {int i,j;for (i=0;string1[i]!='\0';i++)string[i]=string1[i];for(j=0;string2[j]!='\0';j++)string[i+j]=string2[j];string[i+j]='\0';}7-7#include <>int main(){void cpy(char [],char []);char str[80],c[80];printf("input string:");gets(str);cpy(str,c);printf("The vowel letters are:%s\n",c);return 0;}void cpy(char s[],char c[]){ int i,j;for (i=0,j=0;s[i]!='\0';i++)if (s[i]=='a'||s[i]=='A'||s[i]=='e'||s[i]=='E'||s[i]=='i'|| s[i]=='I'||s[i]=='o'||s[i]=='O'||s[i]=='u'||s[i]=='U'){c[j]=s[i];j++;}c[j]='\0';}7-8#include <>#include <>int main(){char str[80];void insert(char []);printf("input four digits:");scanf("%s",str);insert(str);return 0;}void insert(char str[]){int i;for (i=strlen(str);i>0;i--){str[2*i]=str[i];str[2*i-1]=' ';}printf("output:\n%s\n",str);}7-9#include <>int letter,digit,space,others;int main(){void count(char []);char text[80];printf("input string:\n");gets(text);printf("string:");puts(text);letter=0;digit=0;space=0;others=0;count(text);printf("\nletter:%d\ndigit:%d\nspace:%d\nothers:%d\n",letter,digit,space,others );return 0;}void count(char str[]){int i;for (i=0;str[i]!='\0';i++)if ((str[i]>='a'&& str[i]<='z')||(str[i]>='A' && str[i]<='Z'))letter++;else if (str[i]>='0' && str [i]<='9')digit++;else if (str[i]==32)space++;elseothers++;}7-10#include <>#include <>int main(){int alphabetic(char);int longest(char []);int i;char line[100];printf("input one line:\n");gets(line);printf("The longest word is :");for (i=longest(line);alphabetic(line[i]);i++) printf("%c",line[i]);printf("\n");return 0;}int alphabetic(char c){if ((c>='a' && c<='z')||(c>='A'&&c<='z'))return(1);elsereturn(0);}int longest(char string[]){int len=0,i,length=0,flag=1,place=0,point;for (i=0;i<=strlen(string);i++)if (alphabetic(string[i]))if (flag){point=i;flag=0;}elselen++;else{flag=1;if (len>=length){length=len;place=point;len=0;}}return(place);}7-11#include <>#include <>#define N 10char str[N];int main(){void sort(char []);int i,flag;for (flag=1;flag==1;){printf("input string:\n");scanf("%s",&str);if (strlen(str)>N)printf("string too long,input again!"); elseflag=0;}sort(str);printf("string sorted:\n");for (i=0;i<N;i++)printf("%c",str[i]);printf("\n");return 0;}void sort(char str[]){int i,j;char t;for(j=1;j<N;j++)for (i=0;(i<N-j)&&(str[i]!='\0');i++)if(str[i]>str[i+1]){t=str[i];str[i]=str[i+1];str[i+1]=t;}}7-12#include <>#include <>int main(){float solut(float a,float b,float c,float d); float a,b,c,d;printf("input a,b,c,d:");scanf("%f,%f,%f,%f",&a,&b,&c,&d);printf("x=%\n",solut(a,b,c,d));return 0;}float solut(float a,float b,float c,float d) {float x=1,x0,f,f1;do{x0=x;f=((a*x0+b)*x0+c)*x0+d;f1=(3*a*x0+2*b)*x0+c;x=x0-f/f1;}while(fabs(x-x0)>=1e-3);return(x);}7-16#include <>#define MAX 1000int main(){ int htoi(char s[]);int c,i,flag,flag1;char t[MAX];i=0;flag=0;flag1=1;printf("input a HEX number:");while((c=getchar())!='\0' && i<MAX&& flag1){if (c>='0' && c<='9'||c>='a' && c<='f'||c>='A' && c<='F') {flag=1;t[i++]=c;}else if (flag){t[i]='\0';printf("decimal number %d\n",htoi(t));printf("continue or not");c=getchar();if (c=='N'||c=='n')flag1=0;else{flag=0;i=0;printf("\ninput a HEX number:");}}}return 0;}int htoi(char s[]){ int i,n;n=0;for (i=0;s[i]!='\0';i++){if (s[i]>='0'&& s[i]<='9')n=n*16+s[i]-'0';if (s[i]>='a' && s[i]<='f')n=n*16+s[i]-'a'+10;if (s[i]>='A' && s[i]<='F')n=n*16+s[i]-'A'+10;}return(n);}7-17#include <>int main(){ void convert(int n);int number;printf("input an integer: ");scanf("%d",&number);printf("output: ");if (number<0){putchar('-');putchar(' '); /* 先输出一个‘-’号和空格 */ number=-number;}convert(number);printf("\n");return 0;}void convert(int n){ int i;if ((i=n/10)!=0)convert(i);putchar(n%10+'0');putchar(32);}7-18#include <>int main(){int sum_day(int month,int day);int leap(int year);int year,month,day,days;printf("input date(year,month,day):");scanf("%d,%d,%d",&year,&month,&day);printf("%d/%d/%d ",year,month,day);days=sum_day(month,day); /* 调用函数sum_day */ if(leap(year)&&month>=3) /* 调用函数leap */days=days+1;printf("is the %dth day in this year.\n",days);return 0;}int sum_day(int month,int day) /* 函数sum_day:计算日期 */ {int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};int i;for (i=1;i<month;i++)day+=day_tab[i]; /* 累加所在月之前天数 */return(day);} /* 函数leap:判断是否为闰年 */int leap(int year){int leap;leap=year%4==0&&year%100!=0||year%400==0;return(leap);}8-1#include <>int main(){ void swap(int *p1,int *p2);int n1,n2,n3;int *p1,*p2,*p3;printf("input three integer n1,n2,n3:");scanf("%d,%d,%d",&n1,&n2,&n3);p1=&n1;p2=&n2;p3=&n3;if(n1>n2) swap(p1,p2);if(n1>n3) swap(p1,p3);if(n2>n3) swap(p2,p3);printf("Now,the order is:%d,%d,%d\n",n1,n2,n3);return 0;}void swap(int *p1,int *p2){int p;p=*p1; *p1=*p2; *p2=p;}8-2#include <>#include <>int main(){void swap(char *,char *);char str1[20],str2[20],str3[20];printf("input three line:\n");gets(str1);gets(str2);gets(str3);if(strcmp(str1,str2)>0) swap(str1,str2); if(strcmp(str1,str3)>0) swap(str1,str3); if(strcmp(str2,str3)>0) swap(str2,str3); printf("Now,the order is:\n");printf("%s\n%s\n%s\n",str1,str2,str3);return 0;}void swap(char *p1,char *p2){char p[20];strcpy(p,p1);strcpy(p1,p2);strcpy(p2,p); }8-3#include <>int main(){ void input(int *);void max_min_value(int *);void output(int *);int number[10];input(number); max_min_value(number); output(number);return 0;}void input(int *number){int i;printf("input 10 numbers:");for (i=0;i<10;i++)scanf("%d",&number[i]);}void max_min_value(int *number) { int *max,*min,*p,temp;max=min=number;for (p=number+1;p<number+10;p++)if (*p>*max) max=p;else if (*p<*min) min=p;。

c语言程序设计谭浩强课后习题答案

c语言程序设计谭浩强课后习题答案

c语言程序设计谭浩强课后习题答案1. 一个C语言程序是由()。

[单选题] *一个主程序和若干子程序组成函数组成(正确答案)若干过程组成若干子程序组成2. 一个C程序的执行是从()。

[单选题] *本程序的main函数开始,到main函数结束(正确答案)本程序文件的第一个函数开始,到本程序文件的最后一个函数结束本程序的main函数开始,到本程序文件的最后一个函数结束本程序文件的第一个函数开始,到本程序main函数结束3. 同时定义相同类型的多个变量要使用()。

[单选题] *句号分号逗号(正确答案)冒号4. 标准C语言程序编译生成的文件后缀为()。

[单选题] *.c.cpp.obj(正确答案).exe5. 下面的注释那个写法是正确的()。

[单选题] */*this is a comment*\/*this is a comment/* it is error*//*this is a comment*/ (正确答案)/*this one seems like a comment doesn’t it6. 以下叙述中正确的是()。

[单选题] *用C程序实现的算法必须要有输入和输出操作用C程序实现的算法可以没有输出但必须要有输入用C程序实现的算法可以没有输入但必须要有输出(正确答案)用C程序实现的算法可以既没有输入也没有输出7. C语言中3种基本结构是()。

[单选题] *顺序结构、选择结构、循环结构(正确答案)if、switch、breakfor、while、do-whileif、for、continue8. 以下叙述中错误的是()。

[单选题] *用户所定义的标识符允许使用关键字(正确答案)用户所定义的标识符应尽量做到“见名知意”用户所定义的标识符必须以字母或下划线开头用户定义的标识符中,大、小写字母代表不同标识9. 以下不能定义为用户标识符的是()。

[单选题] *Main_0_intsizeof(正确答案)10. 以下选项中合法的用户标识符是()。

谭浩强c程序设计课后习题答案

谭浩强c程序设计课后习题答案
}
double solut(double a,double b,double c,double d)
{double x=1,x0,f,f1;
do
{x0=x;
f=((a*x0+b)*x0+c)*x0+d;
f1=(3*a*x0+2*b)*x0+c;
x=x0-f/f1;
}
while(fabs(x-x0)>=1e-5);
using namespace std;
int main()
{
int a,b,c;
a=10;
b=23;
c=a+b;
cout<<"a+b=";
cout<<c;
cout<<endl;
return 0;
}
七题
#include <iostream>
using namespace std;
int main()
{
int a,b,c;
int f(int x,int y,int z);
cin>>a>>b>>c;
c=f(a,b,c);
cout<<c<<endl;
return 0;
}
int f(int x,int y,int z)
{
int m;
if (x<y) m=x;
else m=y;
if (z<m) m=z;
{int prime(int);
int a,b;
for(a=3;a<=n/2;a=a+2)
{if(prime(a))

C程序设计谭浩强完整课后习题答案

C程序设计谭浩强完整课后习题答案

C程序设计(第四版)(谭浩强)第一章课后习题答案P006 1.1 向屏幕输出文字.#include<stdio.h>//预编译. 代码均调试成功,若有失误大多不是代码问题.自已找找.int main(){printf("Welcome to \n");return 0; //与int main对应,为了程序可移植性,建议全用int main + return 0;.}P008 1.2 求两个数的和.#include<stdio.h>int main(){int a,b,sum;a=5;b=4;sum=a+b;printf("The sum is %d .\n",sum);return 0;}P008 1.3 调用函数比较两个数的大小.#include<stdio.h>int main(){int max(int x,int y); //被调用函数在主函数后面,用前先声明.int a,b,c;scanf("%d,%d",&a,&b); //输入时要按格式来,此处的逗号,用空格会发生错误.c=max(a,b); //a,b作为实参传入被调用函数中.printf("The max is %d .\n",c);return 0;}int max(int x,int y) //定义了两个形参.{int z; //z属于局部变量,可与主函数中相同名字.if (x>y)z=x;elsez=y;return(z); //z作为整个程序的出口值,赋给主函数中的c.}P015 0.6 三个数的大小.(数字0表示课后练习题)#include<stdio.h>int main(){int a,b,c,d; //d是用于存储最大值的.int max(int x , int y , int z); //测试可知,在VS2008中,可以不预先声明.printf("Please input 3 numbers :\n");scanf("%d %d %d",&a,&b,&c);d=max(a,b,c); //调用函数中有三个形参,这里需要传入三个实参,才可运算.printf("The max is :%d .\n",d); // d可以换成max(a,b,c).}int max(int x , int y , int z){int m;if (x>y && x>z) //求三者之大的一种方法.m=x;if (y>x && y>z)m=y;if (z>y && z>x)m=z;return (m); //返回值m给主函数中的d.}C程序设计(第四版)(谭浩强)第2章课后习题答案算法——程序的灵魂#include<stdio.h>int main(){int i,s=1; //在执行数值操作前一定要先有个初值.for(i=1;i<6;i++) //这里是到6.{s=s*i; //相乘}printf("The sum is %d .\n",s);return 0;}#include<stdio.h> //作出要求:换成1到11间奇数相乘.int main(){int i,s=1; //在执行数值操作前一定要先有个初值.for(i=1;i<12;i++) //这里是到,但题目要求的是取单数.也可以是i=i+2{if(i%2!=0) //i对取模,值为非为奇数;为则为偶数.s=s*i;elsecontinue; //跳过这个for循环的这一次,执行下一次.}printf("The sum is %d .\n",s);return 0;}P019 2.2 按要求输出80分以上的学生信息.暂时没法做.P019 2.3 判断2000-2500年中的闰年,并输出.年的概念是地球围绕太阳一周的时间(所谓公转周期)称为一年,这个周期是相当稳定的,很长时间也不会变动1秒,但是真正的一年是365.2423天(目前)。

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using namespace std;
int main()
{int i,j,m,n;
i=8;
j=10;
m=++i+j++;
n=(++i)+(++j)+m;
cout<<i<<'\t'<<j<<'\t'<<m<<'\t'<<n<<endl;
return 0;
}
2.8题
#include <iostream>
int main()
{
int a,b,c;
a=10;
b=23;
c=a+b;
cout<<"a+b=";
cout<<c;
cout<<endl;
return 0;
}
1.7七题
#include <iostream>
using namespace std;
int main()
{
int a,b,c;
int f(int x,int y,int z);
cout<<"s= "<<setw(10)<<s<<endl;
cout<<"sq="<<setw(10)<<sq<<endl;
cout<<"vq="<<setw(10)<<vq<<endl;
cout<<"vz="<<setw(10)<<vz<<endl;
return 0;
}
3.3题
#include <iostream>
cin>>a>>b;
c=add(a,b);
cout<<"a+b="<<c<<endl;
returt y)
{int z;
z=x+y;
return(z);
}
2.3题
#include <iostream>
using namespace std;
int main()
{char c1='a',c2='b',c3='c',c4='\101',c5='\116';
cout<<c1<<c2<<c3<<'\n';
cout<<"\t\b"<<c4<<'\t'<<c5<<'\n';
return 0;
}
2.4题
#include <iostream>
using namespace std;
cin>>r>>h;
l=2*pi*r;
s=r*r*pi;
sq=4*pi*r*r;
vq=3.0/4.0*pi*r*r*r;
vz=pi*r*r*h;
cout<<setiosflags(ios::fixed)<<setiosflags(ios::right)
<<setprecision(2);
cout<<"l= "<<setw(10)<<l<<endl;
int main()
{
int a,b,c;
cin>>a>>b;
c=a+b;
cout<<"a+b="<<a+b<<endl;
return 0;
}
1.9题
#include <iostream>
using namespace std;
int main()
{
int a,b,c;
int add(int x,int y);
return 0;
}
3.2题
#include <iostream>
#include <iomanip>
using namespace std;
int main ( )
{float h,r,l,s,sq,vq,vz;
const float pi=3.1415926;
cout<<"please enter r,h:";
using namespace std;
int main()
{char c1='C', c2='h', c3='i', c4='n', c5='a';
c1+=4;
c2+=4;
c3+=4;
c4+=4;
c5+=4;
cout<<"password is:"<<c1<<c2<<c3<<c4<<c5<<endl;
cin>>a>>b>>c;
c=f(a,b,c);
cout<<c<<endl;
return 0;
}
int f(int x,int y,int z)
{
int m;
if (x<y) m=x;
else m=y;
if (z<m) m=z;
return(m);
}
1.8题
#include <iostream>
using namespace std;
using namespace std;
int main ( )
{char c1,c2;
cout<<"请输入两个字符c1,c2:";
c1=getchar(); //将输入的第一个字符赋给c1
c2=getchar(); //将输入的第二个字符赋给c2
cout<<"用putchar函数输出结果为:";
putchar(c1);
putchar(c2);
cout<<endl;
cout<<"用cout语句输出结果为:";
cout<<c1<<c2<<endl;
return 0;
}
3.4题另一解
#include <iostream>
using namespace std;
int main ( )
{char c1,c2;
cout<<"请输入两个字符c1,c2:";
using namespace std;
int main ()
{float c,f;
cout<<"请输入一个华氏温度:";
cin>>f;
c=(5.0/9.0)*(f-32); //注意5和9要用实型表示,否则5/9值为0
cout<<"摄氏温度为:"<<c<<endl;
return 0;
};
3.4题
#include <iostream>
c1=getchar(); //将输入的第一个字符赋给c1
c2=getchar(); //将输入的第二个字符赋给c2
cout<<"用putchar函数输出结果为:";
第一章
1.5题
#include <iostream>
using namespace std;
int main()
{
cout<<"This"<<"is";
cout<<"a"<<"C++";
cout<<"program.";
return 0;
1.6题
#include <iostream>
using namespace std;
int main()
{char c1='C',c2='+',c3='+';
cout<<"I say: \""<<c1<<c2<<c3<<'\"';
cout<<"\t\t"<<"He says: \"C++ is very interesting!\""<< '\n';
return 0;
}
2.7题
#include <iostream>
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