C++实现简单24点游戏

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

随机生成4个代表扑克牌牌面的数字字母，程序自动列出所有可能算出24的表达式，用擅长的语言（C/C++/Java或其他均可）实现程序解决问题。

程序风格良好(使用自定义注释模板)

列出表达式无重复。

以下为源代码

#include
#include
#include  
#include
using namespace std;
char card[] = { 'A','2','3','4','5','6','7','8','9','10','J','Q','K' };
char buf[4];
double nums[4];
char ope[4] = { '+','-','*','/' };
 
 
void cre()//生成
{
 int i = 0;
 int j;
 cout << "生成的四张牌面为：";
 srand((unsigned)time(0));
 for (i = 0; i<4; i++)
 {
 j =rand() % 13;
 buf[i] = card[j];
 }
 cout << buf[0] << ";" << buf[1] << ";" << buf[2] << ";" << buf[3] << "。" << endl;
 for (i = 0; i<4; i++)
 {
 if (buf[i] == 'A') nums[i] = 1;
 else if(buf[i] == '2') nums[i] = 2;
 else if (buf[i] == '3') nums[i] = 3;
 else if (buf[i] == '4') nums[i] = 4;
 else if (buf[i] == '5') nums[i] = 5;
 else if (buf[i] == '6') nums[i] = 6;
 else if (buf[i] == '7') nums[i] = 7;
 else if (buf[i] == '8') nums[i] = 8;
 else if (buf[i] == '9') nums[i] = 9;
 else if (buf[i] == '10') nums[i] = 10;
 else if (buf[i] == 'J') nums[i] = 11;
 else if (buf[i] == 'Q') nums[i] = 12; 
 else if (buf[i] == 'K') nums[i] = 13;
 }
}
 
double calcute(double a, double b, char index)
{
 if (index == '+') return a + b; //若为+，则返回相应结果 
 else if (index == '-') return a - b;
 else if (index == '*') return a*b;
 else if (index == '/')
 if (b != 0)
 return a / b; //只有当分母不为0时，返回结果 
}
 
void exh()//穷举计算
{
 double temp[3], tem[2]; //第一个符号放置后，经过计算后相当于剩下三个数，这个数组用于存储这三个数 
 double sum; //求得的和 
 int judge = 0; //判断是否找到一个合理的解 
 for (int i = 0; i < 4; i++) //第一次放置的符号 
 {
 for (int j = 0; j < 4; j++) //第二次放置的符号 
 {
 for (int k = 0; k < 4; k++) //第三次放置的符号 
 {
 for (int m = 0; m < 3; m++)  //首先计算的两个相邻数字，共有3种情况，相当于括号的作用  
 {
  if (nums[m + 1] == 0 && ope[i] == '/') break;
  temp[m] = calcute(nums[m], nums[m + 1], ope[i]);
  temp[(m + 1) % 3] = nums[(m + 2) % 4];
  temp[(m + 2) % 3] = nums[(m + 3) % 4];  //先确定首先计算的两个数字，计算完成相当于剩下三个数，按顺序储存在temp数组中  
  for (int n = 0; n < 2; n++)  //三个数字选出先计算的两个相邻数字，两种情况，相当于第二个括号  
  {
  if (temp[n + 1] == 0 && ope[j] == '/') break;
  tem[n] = calcute(temp[n], temp[n + 1], ope[j]);
  tem[(n + 1) % 2] = temp[(n + 2) % 3];  //先确定首先计算的两个数字，计算完成相当于剩下两个数，按顺序储存在temp数组中  
  if (tem[1] == 0 && ope[k] == '/') break;
  sum = calcute(tem[0], tem[1], ope[k]);  //计算和    
  if (sum == 24)  //若和为24  
  {
  judge = 1;   //判断符为1，表示已求得解   
  if (m == 0 && n == 0)
  cout << "((" << nums[0] << ope[i] << nums[1] << ")" << ope[j] << nums[2] << ")" << ope[k] << nums[3] << "=" << sum << endl;
  else if (m == 0 && n == 1)
  cout << "(" << nums[0] << ope[i] << nums[1] << ")" << ope[k] << "(" << nums[2] << ope[j] << nums[3] << ")=" << sum << endl;
  else if (m == 1 && n == 0)
  cout << "(" << nums[0] << ope[j] << "(" << nums[1] << ope[i] << nums[2] << ")" << ope[k] << nums[3] << "=" << sum << endl;
  else if (m == 1 && n == 1)
  cout << nums[0] << ope[k] << "((" << nums[1] << ope[i] << nums[2] << ")" << ope[j] << nums[3] << ")=" << sum << endl;
  else if (m == 2 && n == 0)
  cout << "(" << nums[0] << ope[j] << nums[1] << ")" << ope[k] << "(" << nums[2] << ope[i] << nums[3] << ")=" << sum << endl;
  else if (m == 2 && n == 0)
  cout << nums[0] << ope[k] << "(" << nums[1] << ope[j] << "(" << nums[2] << ope[i] << nums[3] << "))=" << sum << endl;   //m=0,1,2 n=0,1表示六种括号放置可能，并按照这六种可能输出相应的格式的计算式    
  }
  }
 }
 }
 }
 }
 if (judge == 0)
 cout << "这四张扑克牌无法找到一个合理的解" << endl; //如果没有找到结果，符号位为0 
}
 
 
 
int main()
{
 int i;
 int select = 1;
 cout<< " ################################################" << endl
 << " #            #" << endl
 << " #    欢迎进入24点游戏    #" << endl
 << " #            #" << endl
 << " ################################################" << endl;
 
 while (select)
 {
 cout<< " ################################################" << endl
 << " #            #" << endl
 << " #    是否开始游戏     #" << endl
 << " #            #" << endl
 << " #   0.是    1.否    #" << endl
 << " #            #" << endl
 << " ################################################" << endl;
 cout << "请输入你的选择（0或1）：";
 cin >> i;
 switch (i)
 {
 case 0:
 cre();
 exh();
 break;
 case 1:
 select = 0;
 break;
 default:
 cout << "请在0和1之间选择！" << endl;
 }
 }
 return 0;
}

效果图1

效果图2

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脚本之家。