算法设计程序源码

1.主元素#include <stdio.h>#include <stdlib.h>int candidate(int m,int array[],int n) {int c = array[m];int count=1;int j=m;while((j<n)&&(count>0)){j++;if (array[j]==c){count++;}else{count--;}}if (j==n){return c;}else{return candidate(j+1,array,n);}}int Majority(int array[],int n){int c=candidate(1,array,n);int count=0;int j=0;while (j<n){if (c==array[j]){count++;}j++;}if (count>(n/2)){return c;}else{return -1;}}int main(){int array[10]={2,4,4,5,4,7,4,9,4,4};int num=0;num=Majority(array,10);printf("%d\n",num);return 0;}2.圈乘运算#include<stdio.h>#include<stdlib.h>#include<math.h>#define MAX 2000000000int A[1650][6];void factor(int x,int T[]){int min=10,max=0,sum=0,t;while(x){t=x%10;x=x/10;if(t>max)max=t;if(t<min)min=t;sum+=t;}T[0]=MAX;T[1]=sum;T[2]=max;T[3]=min;T[4]=0;T[5]=0;}int Compute_Mul_Times(int x,int k) {int i,j,y,s,flag;if(x==0&&k==0)return -1;y=(int)log10((float)x)+1;y=y*81+9;if(y<171)y=171;if(k>y){return -1;}if(k==x) {return 0;}for(i=1;i<=y;i++){factor(i,A[i]);}if (x>y){y=y+1;factor(x,A[y]);A[y][0]=0;}else{A[x][0]=0;}flag=1;while(flag){flag=0;for(i=1;i<=y;i++){if(A[i][0]<MAX){for(j=1;j<=y;j++){if(A[j][0]<MAX){s=A[i][1]*A[j][2]+A[j][3];if(A[i][0]+A[j][0]+1<A[s][0]){A[s][0]=A[i][0]+A[j][0]+1;A[s][4]=i;A[s][5]=j;flag=1;}}}}}}if(A[k][0]<MAX){// printf("最少需要圈乘次数：%d\n 最后圈乘的数字：%d,%d\n",A[k][0], A[k][4], A[k][5]);return A[k][0];}else{return -1;}}void main() {int x,k,time;printf("请输入数字：");scanf("%d%d",&x,&k);if((time=Compute_Mul_Times(x,k))!=-1) {printf("最少需要圈乘次数：%d\n",time);}elseprintf("无解！");}3.删数最小值问题#include<stdio.h>#define M 240main(){char ch[M]; /*数串*/int r[M]; /*余下的数符在原数中的位置*/int l; /*数串的实际长度*/int s; /*删除的数符的个数*/int i;int start_pos=0,pos=-1;char min='A';/*数据输入和数据的初始化*/printf("\n***********删数问题***************\n");printf("\n输入数串N=");gets(ch);printf("删除的数符个数S=");scanf("%d",&s);l=0;for(i=0;ch[i]!='\0';i++){r[i]=i;l++;}while (s>0){min='A';for (i=start_pos;i<=start_pos+s;i++){if (min>ch[i]){min=ch[i];pos=i;}}for (i=start_pos;i<pos;i++){r[i]=-1;s--;}start_pos=pos+1;}printf("删数结果为：");for (i=0;i<l;i++){if (r[i]!=-1){printf("%c",ch[r[i]]);}}printf("\n");}4.独立钻石跳棋#include <iostream>#include <fstream>using namespace std;struct step //记录移动棋子的信息{int sx, sy; //记录移动棋子前棋子的位置int tx, ty; //记录移动棋子后棋子的位置int dir; // dir值代表移动棋子的方向};struct step mystack[100], last_step;char diamond[7][7];int Left_diamond = 32;int x, y, nx, ny, ndir, top; //ndir值代表方向, 0代表向右, 1代表向下, 2代表向左, 3代表向上int flag=1; // 是否成功找到解的标志void Init_diamond(){for(int i=0; i<7; i++){for(int j=0; j<7; j++){if((i<2 || i>4) && (j<2 || j>4));else{diamond[i][j] = '*';}}}diamond[3][3] = '!';}int Move_diamond(int y, int x, int dir){if(diamond[y][x] != '*'){return 0;}struct step temp;switch(dir){case 0:if(x+2>6 || diamond[y][x+1]!='*' || diamond[y][x+2]!='!'){return 0;}diamond[y][x] = diamond[y][x+1] = '!';diamond[y][x+2] = '*';temp.sx = x;temp.sy = y;temp.tx = x+2;temp.ty = y;temp.dir = dir;mystack[top++] = temp;return 1;break;case 1:if(y+2>6 || diamond[y+1][x]!='*' || diamond[y+2][x]!='!'){return 0;}diamond[y][x] = diamond[y+1][x] = '!';diamond[y+2][x] = '*';temp.sx = x;temp.sy = y;temp.tx = x;temp.ty = y+2;temp.dir = dir;mystack[top++] = temp;return 1;break;case 2:if(x-2<0 || diamond[y][x-1]!='*' || diamond[y][x-2]!='!'){return 0;}diamond[y][x] = diamond[y][x-1] = '!';diamond[y][x-2] = '*';temp.sx = x;temp.sy = y;temp.tx = x-2;temp.ty = y;temp.dir = dir;mystack[top++] = temp;return 1;break;case 3:if(y-2<0 || diamond[y-1][x]!='*' || diamond[y-2][x]!='!'){return 0;}diamond[y][x] = diamond[y-1][x] = '!';diamond[y-2][x] = '*';temp.sx = x;temp.sy = y;temp.tx = x;temp.ty = y-2;temp.dir = dir;mystack[top++] = temp;return 1;break;default:break;}return 0;}void main(){// 输出一个解到文本文件result.txtofstream answer("result.txt");Init_diamond();top = nx = ny = ndir = 0;// 回溯遍历，直到找到一个解while(1){if(Left_diamond == 1 && diamond[3][3] == '*'){break;}for(y=ny; y<7; y++,nx=0){for(x=nx; x<7; x++,ndir=0){for(int dir=ndir; dir<4; dir++){if(Move_diamond(y, x, dir)){Left_diamond--;nx = ny = ndir = 0;goto nextstep;}}}}nextstep:if(y == 7){top--;// 回到上一步，并改变方向if(top >= 0){last_step = mystack[top];diamond[(last_step.sy + last_step.ty)/2][(last_step.sx + last_step.tx)/2] = '*';diamond[last_step.sy][last_step.sx] = '*';diamond[last_step.ty][last_step.tx] = '!';nx = last_step.sx;ny = last_step.sy;ndir = last_step.dir + 1;Left_diamond++;}else{answer<<"无解！"<<endl;cout<<"无解！"<<endl;flag=0;break;}}}Init_diamond();answer<<"移动解如下："<<endl;for(int i=0; i<7; i++){for(int j=0; j<7; j++){answer<<diamond[i][j]<<' ';}answer<<endl;}answer<<endl<<endl;// 输出步骤cout<<"棋子跳动的过程如下！"<<"\n";for(int n=0; n<top; n++){cout<< "step "<<n+1<<":"<<"\n" "Move diamond ("<<mystack[n].sy+1<<","<< mystack[n].sx+1<<") --->("<<mystack[n].ty+1<<","<<mystack[n].tx+1 <<")"<<endl;diamond[mystack[n].sy][mystack[n].sx] = '!';diamond[(mystack[n].sy+mystack[n].ty)/ 2][(mystack[n].sx+mystack[n].tx)/2] = '!';diamond[mystack[n].ty][mystack[n].tx] = '*';cout<< "Left diamonds : "<<top-n<<endl;for(int k=0; k<7; k++){for(int j=0; j<7; j++){answer<<diamond[k][j]<<' ';}answer<<endl;}answer<<endl<<endl;}if(flag){cout<<"移动解已经输出到\"result.txt\" 文件中！"<<endl;}}5.背包问题#include<iostream>#include<stack>using namespace std;#define N 100class HeapNode //定义HeapNode结点类{public:double upper,price,weight; //upper为结点的价值上界，price是结点所对应的价值，weight为结点所相应的重量int level,x[N]; //活节点在子集树中所处的层序号};double MaxBound(int i);double Knap();void AddLiveNode(double up,double cp,double cw,bool ch,int level);stack<HeapNode> High; //最大队High double w[N],p[N]; //把物品重量和价值定义为双精度浮点数double cw,cp,c=30; //cw为当前重量，cp 为当前价值，定义背包容量为30int n=3; //货物数量为3int main(){cout<<"请按顺序输入3个物品的重量：（按回车键区分每个物品的重量）"<<endl;int i;for(i=1;i<=n;i++)cin>>w[i]; //输入3个物品的重量cout<<"请按顺序输入3个物品的价值：（按回车键区分每个物品的价值）"<<endl;for(i=1;i<=n;i++)cin>>p[i]; //输入3个物品的价值cout<<"最大价值为：";cout<<Knap()<<endl; //调用knap函数输出最大价值return 0;}double MaxBound(int j) //MaxBound函数求最大上界{double left=c-cw,b=cp; //剩余容量和价值上界while(j<=n&&w[j]<=left) //以物品单位重量价值递减装填剩余容量{left-=w[j];b+=p[j];j++;}if(j<=n)b+=p[j]/w[j]*left; //装填剩余容量装满背包return b;}void AddLiveNode(double up,double cp,double cw,bool ch,int lev)//将一个新的活结点插入到子集数和最大堆High中{HeapNode be;be.upper=up;be.price=cp;be.weight=cw;be.level=lev;if(lev<=n)High.push(be); //调用stack头文件的push函数}double Knap() //优先队列分支限界法，返回最大价值，bestx返回最优解{int i=1; cw=cp=0; doublebestp=0,up=MaxBound(1); //调用MaxBound求出价值上界，best为最优值while(1) //非叶子结点{double wt=cw+w[i];if(wt<=c) //左儿子结点为可行结点{if(cp+p[i]>bestp)bestp=cp+p[i];AddLiveNode(up,cp+p[i],cw+w[i],true,i +1);}up=MaxBound(i+1);if(up>=bestp) //右子数可能含最优解AddLiveNode(up,cp,cw,false,i+1);if(High.empty()) return bestp;HeapNode node=High.top(); //取下一扩展结点High.pop();cw=node.weight; cp=node.price; up=node.upper; i=node.level;}}。