C语言实现24点游戏源代码
时间：2020-12-05 13:08:36|栏目：C代码|点击： 次
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 50
double 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 0
int 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))*8 6/((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))*6 Total[13], Success andy@ubuntu14:~/work$ ./test 7 7 7 7 Fail
