本文实例为大家分享了C语言实现24点游戏的具体代码，供大家参考，具体内容如下

参考文章：C语言实现经典24点算法

将算法实现改成C语言，并可在linux服务器上运行。同时修改为可显示所有结果。
注：如果传参重复，如4，4，7，7这样，会回显重复结果，暂无法清除。

#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;}

运行结果如下：

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

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。