习题09答案

习题9答案1．解答1：#include <stdio.h>void swap( int *pa, int *pb, int *pc){int t;if(*pa>*pb){ t=*pa; *pa=*pb; *pb=t; }if(*pa>*pc){ t=*pa; *pa=*pc; *pc=t; }if(*pb>*pc){ t=*pb; *pb=*pc; *pc=t; }}main( ){int a, b, c;scanf("%d%d%d", &a, &b, &c);swap(&a,&b,&c);printf("%d\t%d\t%d\n", a, b, c);} 解答2：#include <stdio.h>void swap2( int *m, int *n){int t;t=*m; *m=*n; *n=t;}void swap( int *pa, int *pb, int *pc) {if(*pa>*pb) swap2(pa, pb);if(*pa>*pc) swap2(pa, pc);if(*pb>*pc) swap2(pb, pc);}main( ){int a, b, c;scanf("%d%d%d", &a, &b, &c);swap(&a,&b,&c);printf("%d\t%d\t%d\n", a, b, c); }2．解答：#include <stdio.h>void stat(char *str, int *letter, int *digit, int *space, int *other){char c;for( ; *str; str++){c=*str;if(('a'<=c&&c<='z')||('A'<=c&&c<='Z'))(*letter)++;else if(c==' ')(*space)++;else if('0'<=c && c<='9')(*digit)++;else(*other)++;}}main( ){char str[100];int letter=0, digit=0, space=0, other=0; /* 变量letter用于统计字母的个数*//* 变量digit用于统计数字字符的个数*//* 变量space用于统计空格的个数*//* 变量other用于统计其他字符的个数*/gets(str);stat(str, &letter, &digit, &space, &other);printf("letter=%d\n", letter);printf(" space=%d\n", space);printf(" digit=%d\n", digit);printf(" other=%d\n", other);}程序的一次运行过程如下：She has 5 apples.letter=12space=3digit=1other=13．解答：#include <stdio.h>void output(int *p, int n){int i;for(i=0; i<n; i++)printf("%d\t", *p++);printf("\n");}void change(int a[10], int n){int t,*pend,*p,*maxp,*minp;minp=a;pend=a+n;for(p=a+1; p<pend; p++)if(*p<*minp) minp=p;if(minp!=a){t=*minp; *minp=*a; *a=t;}maxp=a;for(p=a+1;p<pend;p++)if(*p>*maxp) maxp=p;if(maxp!=a+n-1){t=*maxp; *maxp=*(a+n-1); *(a+n-1)=t;}}main (){int a[10],n,i;scanf("%d", &n);for(i=0; i<n; i++)scanf("%d", (a+i));output(a,n); /* 输出原数组*/change(a,n);output(a,n); /* 输出调换后的数组*/}4．解答1：算法：将数组a的最低k位存入一临时数组temp中，将数组a的高n-k位平移至数组a最低n-k位，然后将临时数组temp中的k位复制回数组a的高k位。

循环右移k位，相当于如下操作：k位n-k位#include <stdio.h>void output(int *p, int n){int i;for(i=0; i<n; i++)printf("%d\t", *p++);printf("\n");}void right_move(int a[10], int n, int k){int *pa1, *pa2, *tp, temp[10], i;tp=temp;pa1=a+n-k;for(i=0; i<k; i++) /* 将数组a右侧k个数复制到临时数组temp中*/ {*tp=*pa1;tp++;pa1++;}pa1=a+n-1;pa2=a+n-1-k;for(i=0; i<n-k; i++) /* 将数组a前n-k个数移到数组a最右侧n-k位*/ {*pa1 = *pa2;pa1--; pa2--;}tp--;for(i=0; i<k; i++) /* 将临时数组temp中的k个元素复制到数组a最前面k位*/{*pa1 = *tp;pa1--; tp--;}tp++;}main (){int a[10],k,i;printf("Please input 10 elements: ");for(i=0; i<10; i++)scanf("%d", a+i);printf("Please input k: ");scanf("%d", &k);if(k<0)k=10+k; /* 转换成右移*/output(a,10); /* 输出原数组*/right_move(a,10, k);output(a,10); /* 输出调换后的数组*/}解答2：算法：函数move1()实现将数组循环右移一位，函数right_move()调用k次move1()实现将数组循环右移k位。

#include <stdio.h>void output(int *p, int n){int i;for(i=0; i<n; i++)printf("%d\t", *p++);printf("\n");}void move1(int a[], int n) /* 循环右移一位*/{int i, t;if(n==0) return;t=a[n-1];for(i=n-1; i>0; i--)a[i]=a[i-1];a[0]=t;}void right_move(int a[10], int n, int k){int i;for(i=0; i<k; i++)move1(a,n); /* 调用move1()将a数组右移k位*/ }main () {int a[10],k,i;printf("Please input 10 elements: ");for(i=0; i<10; i++)scanf("%d", a+i);printf("Please input k: ");scanf("%d", &k);if(k<0)k=10+k; /* 转换成右移*/ output(a,10); /* 输出原数组*/right_move(a,10, k);output(a,10); /* 输出调换后的数组*/ }5．解答：#include <stdio.h>void select_odd_index(char *str1, char *str2) {str1++;while(*str1){*str2 = *str1;str2++;str1++;if(*str1=='\0')break;str1++;}*str2='\0';}main ( ){char str1[100],str2[50];printf("Please input a string: ");gets(str1);select_odd_index(str1, str2);puts(str2);}6．解答：#include <stdio.h>#include <string.h>void select_max(char *p1, char *p2) {char temp[100];int maxlen=0,len;while(*p1){while(*p1==' ') p1++;len=0;while( *p1!=' ' && *p1!='\0' ){*(temp+len)=*p1;len++;p1++;}*(temp+len)='\0';if(len>maxlen){maxlen=len;strcpy(p2,temp);}}}main (){char main_str[100],sub_str[50];printf("Please input main string: ");gets(main_str);select_max(main_str, sub_str);puts(sub_str);}7．解答：#include <stdio.h>int find_saddle_point(int a[ ][5], int *rowp, int *colp){int i, j, flag, max, maxj, k;for(i=0; i<4; i++){max=a[i][0];maxj=0;for(j=1; j<5; j++) /* 找出第i行中最大值，并记下该最大值所在的列号*/if(a[i][j]>max){*rowp=i;*colp=j;max=a[i][j]; maxj=j;}flag=1;for(k=0; k<4; k++) /* 判断是否为该列中最小值*/if(a[k][maxj]<max){flag=0;break;}if(flag) /* 找到鞍点，输出行号列号及鞍点值*/return(1);}return(0);}main ( ){int a[4][5]={{2,6,3,4,7},{5,6,5,5,1},{5,7,6,7,3},{1,9,2,7,5}}, i, j;for(i=0; i<4; i++){for(j=0;j<5;j++)printf("%d\t", a[i][j]);printf("\n");}printf("\n");if(find_saddle_point(a,&i,&j))printf("saddle point is a[%d][%d]\n", i, j);elseprintf("Can't found!\n");}8．解答：#include <stdio.h>#include <string.h>#include <ctype.h>void filter(char s1[]) /* 过滤: 去除非字母，并全部转换成大写*/{char s2[80],*p1=s1,*p2=s2;while(*p1)if(isalpha(*p1)) /* isalpha( )参数若是字母返回真*/{*p2=toupper(*p1); /* toupper( )函数将参数字母转换成大写字母*/p1++; p2++;}else p1++;*p2=0;strcpy(s1,s2);}int palin(char s[]){char *head,*tail;char s1[80]; /* 临时字符串*/strcpy(s1,s);filter(s1);head=s1;tail=s1+strlen(s1)-1;while(head<tail)if(*head==*tail) head++,tail--;else return(0);return(1);}void main(){char s[80];gets(s);if(palin(s))printf("Yes.\n");elseprintf("No.\n");}palin( ) 函数另解1：int palin(char s[]){char *head,*tail;head=s;tail=s+strlen(s)-1;while(head<tail){while(!isalpha(*head)) head++;while(!isalpha(*tail)) tail--;if(toupper(*head)==toupper(*tail))head++,tail--;else return(0);}return(1);}palin( ) 函数另解2：int palin(char s[]){char *p1=s,*p2=s+strlen(s)-1;strupr(s); /* strupr(s) 是字符串处理函数，功能是将字符串中全部字符转换成大写*/while(p1<p2)if('A'<=*p1 && *p1<='Z'){if('A'<=*p2 && *p2<='Z'){if(*p1!=*p2)return(0);elsep1++,p2--;}elsep2--;}elsep1++;return(1);}9．解答：#include <stdio.h>int my_strcmp(char *s1, char *s2) {while(*s1==*s2){if(*s1==0) return 0;s1++;s2++;}return(*s1-*s2);}main( ){char a[80], b[80];int result;printf("Please input the first string: ");gets(a);printf("Please input the second string: ");gets(b);result = my_strcmp(a, b);if(result>0)printf("s1>s2, result=%d\n", result);else if(result==0)printf("s1==s2\n");elseprintf("s1<s2, result=%d\n", result); }10．解答：#include <stdio.h>#include <string.h>void swap2(char *s1, char *s2){char temp[80];strcpy(temp,s1);strcpy(s1,s2);strcpy(s2,temp);}void swap(char *s1, char *s2, char *s3){if(strcmp(s1,s2)>0) swap2(s1,s2);if(strcmp(s1,s3)>0) swap2(s1,s3);if(strcmp(s2,s3)>0) swap2(s2,s3);}main( ){char str1[80], str2[80],str3[80];gets(str1);gets(str2);gets(str3);swap(str1,str2,str3);printf("%s,%s,%s\n", str1, str2, str3);}11．解答：#include <stdio.h>main( ){int month;char *p[12]={"January","February","March","April","May","June", "July","August","September","October","November","December"};printf("Please input month: ");scanf("%d", &month);while(0<month && month<13){printf("%s\n", p[month-1]);printf("Please input month: ");scanf("%d", &month);}}12．解答1：#include <stdio.h>int selectnum(char *ps, int *pb) /* ps指向字符串，pb指向整型数组*/ {int count=0; /* 整数串个数*/while(*ps!='\0'){if (*ps>='0'&&*ps<='9')/* 碰到数字串的第一个字符*/{*pb = *pb*10 + *ps - '0';if(*(ps+1)<'0'||*(ps+1)>'9')pb++,count++;}ps++;}return count;}main (){char s[100];int b[10]={0},count,i;printf("please input a string: \n");gets(s); /* 输入带空格的字符串*/count = selectnum(s, b);printf("count=%d\n", count);for (i=0; i<count; i++)printf("b[%d]=%d\n", i, *(b+i));}解答2：#include <stdio.h>int selectnum(char *ps, int *pb) /* ps指向字符串，pb指向整型数组*/ {int count; /* 整数串个数*/for(count=0; *ps!='\0'; ){while(*ps!='\0' && (*ps<'0' || *ps>'9') ) /* 跳过非数字串*/ps++;if(*ps>='0'&&*ps<='9') count++;for(; *ps>='0'&&*ps<='9'; ps++)/* 处理数字串*/pb[count-1]=pb[count-1]*10+*ps-'0';}return count;}main (){char s[100];int b[10]={0},count,i;printf("please input a string: \n");gets(s); /* 输入带空格的字符串*/count = selectnum(s, b);printf("count=%d\n", count);for (i=0; i<count; i++)printf("b[%d]=%d\n", i, *(b+i));}13．解答：#include <stdio.h>void count(int *a, int n, int m){int *pa, *p, total=n, i;pa=a;while(total){for(i=1; i<m; i++) /* 指针向右移动m-1次*/{pa++;if(pa==a+total) pa=a;}printf("%d%c", *pa, (total==1)?' ':','); /* 找到并输出当前应退出的人的序号*/if(pa==a+total-1) pa=a; /* 如果当前退出的人恰好是数组最右一个元素*/else{p=pa;while(p<a+total-1) /* 当前元素将后面的元素依次左移一个元素*/{*p=*(p+1);p++;}}total--;}printf("\n");}main (){int n, m, a[100], i;printf("Please input n & m: ");scanf("%d%d", &n, &m);for(i=1; i<=n; i++)a[i-1]=i;count(a, n, m);}。