C语言实现24点游戏源代码

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h> const double PRECISION = 1E-6; #define COUNT 4 const int RESULT = 24; #define STRLEN 50double number[COUNT] = {0};  //这里一定要用double,char expression[COUNT][STRLEN] = {0}; //保存表达式 #define TRUE 1#define FALSE 0int cnt = 0;void Test(int n){  int i = 0; int j = 0; int len = 0; //递归结束  if(1 == n){   if(number[0] == RESULT)  {    // 避免输出前后括号    for (i = 1; i < strlen(expression[0]) - 1; i++)    {     printf("%c", expression[0][i]);    }    printf("/n");    cnt++;   return;   }   else    return;  }  //递归过程  for(i=0;i<n;i++){   for(j=i+1;j<n;j++){    double a,b;    char expa[STRLEN] = {0};   char expb[STRLEN] = {0};   a=number[i];    b=number[j];    // 删除number[j]元素，用number[n-1]填补    number[j]=number[n-1];    strcpy(expa, expression[i]);   strcpy(expb, expression[j]);   // 删除expression[j]元素，用expression[n-1]填补    strcpy(expression[j], expression[n-1]);   // 加法    len= strlen(expression[i]);   snprintf(expression[i], STRLEN, "(%s+%s)", expa, expb);   number[i]=a+b;    Test(n-1);   //减号有两种情况,a-b与b-a    len= strlen(expression[i]);   snprintf(expression[i], STRLEN, "(%s-%s)", expa, expb);   number[i]=a-b;    Test(n-1);    if(a != b)   {    len= strlen(expression[i]);    snprintf(expression[i], STRLEN, "(%s-%s)", expb, expa);     number[i]=b-a;     Test(n-1);    }   // 乘法    len= strlen(expression[i]);   snprintf(expression[i], STRLEN, "(%s*%s)", expa, expb);    number[i]=a*b;    Test(n-1);    //除法也有两种情况,a/b与b/a    if(b!=0){     len= strlen(expression[i]);    snprintf(expression[i], STRLEN, "(%s/%s)", expa, expb);    number[i]=a/b;     Test(n-1);   }    if((a!=0) && (a != b)){     len= strlen(expression[i]);    snprintf(expression[i], STRLEN, "(%s/%s)", expb, expa);    number[i]=b/a;     Test(n-1);    }    //恢复数组    number[i]=a;    number[j]=b;    strcpy(expression[i], expa);   strcpy(expression[j], expb);  }  }  return; } int main(int argc, char **argv){  int i = 0; if(5 != argc) {  printf("arg err/n");  return 0; } for(i=0;i<COUNT;i++) {   char buffer[20];   number[i] = atoi(argv[i + 1]);  strcpy(expression[i], argv[i + 1]); }  Test(COUNT); if(0 != cnt)  {  printf("Total[%d], Success/n", cnt);  } else  {  printf("Fail/n");  } return 0;} 

andy@ubuntu14:~/work$ ./test 5 6 7 8((5+7)-8)*6(5+7)*(8-6)8/((7-5)/6)(6/(7-5))*86/((7-5)/8)(8/(7-5))*6(6*8)/(7-5)((5-8)+7)*6(7-(8-5))*6(5+7)*(8-6)(6*8)/(7-5)(5+(7-8))*6(5-(8-7))*6Total[13], Successandy@ubuntu14:~/work$ ./test 7 7 7 7Fail