C++實現(xiàn)簡單24點游戲

更新時間：2020年03月06日 13:10:58 作者：JaCo Wu

這篇文章主要為大家詳細介紹了C++實現(xiàn)簡單24點游戲，文中示例代碼介紹的非常詳細，具有一定的參考價值，感興趣的小伙伴們可以參考一下

本文實例為大家分享了C++實現(xiàn)簡單24點游戲的具體代碼，供大家參考，具體內容如下

隨機生成4個代表撲克牌牌面的數(shù)字字母，程序自動列出所有可能算出24的表達式，用擅長的語言（C/C++/Java或其他均可）實現(xiàn)程序解決問題。

程序風格良好(使用自定義注釋模板)

列出表達式無重復。

以下為源代碼

#include<iostream>
#include<string>
#include <stdlib.h> 
#include<time.h>
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 << "生成的四張牌面為：";
 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; //若為+，則返回相應結果 
 else if (index == '-') return a - b;
 else if (index == '*') return a*b;
 else if (index == '/')
 if (b != 0)
 return a / b; //只有當分母不為0時，返回結果 
}
 
void exh()//窮舉計算
{
 double temp[3], tem[2]; //第一個符號放置后，經過計算后相當于剩下三個數(shù)，這個數(shù)組用于存儲這三個數(shù) 
 double sum; //求得的和 
 int judge = 0; //判斷是否找到一個合理的解 
 for (int i = 0; i < 4; i++) //第一次放置的符號 
 {
 for (int j = 0; j < 4; j++) //第二次放置的符號 
 {
 for (int k = 0; k < 4; k++) //第三次放置的符號 
 {
 for (int m = 0; m < 3; m++)  //首先計算的兩個相鄰數(shù)字，共有3種情況，相當于括號的作用  
 {
  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];  //先確定首先計算的兩個數(shù)字，計算完成相當于剩下三個數(shù)，按順序儲存在temp數(shù)組中  
  for (int n = 0; n < 2; n++)  //三個數(shù)字選出先計算的兩個相鄰數(shù)字，兩種情況，相當于第二個括號  
  {
  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];  //先確定首先計算的兩個數(shù)字，計算完成相當于剩下兩個數(shù)，按順序儲存在temp數(shù)組中  
  if (tem[1] == 0 && ope[k] == '/') break;
  sum = calcute(tem[0], tem[1], ope[k]);  //計算和    
  if (sum == 24)  //若和為24  
  {
  judge = 1;   //判斷符為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表示六種括號放置可能，并按照這六種可能輸出相應的格式的計算式    
  }
  }
 }
 }
 }
 }
 if (judge == 0)
 cout << "這四張撲克牌無法找到一個合理的解" << endl; //如果沒有找到結果，符號位為0 
}
 
 
 
int main()
{
 int i;
 int select = 1;
 cout<< " ################################################" << endl
 << " #            #" << endl
 << " #    歡迎進入24點游戲    #" << endl
 << " #            #" << endl
 << " ################################################" << endl;
 
 while (select)
 {
 cout<< " ################################################" << endl
 << " #            #" << endl
 << " #    是否開始游戲     #" << endl
 << " #            #" << endl
 << " #   0.是    1.否    #" << endl
 << " #            #" << endl
 << " ################################################" << endl;
 cout << "請輸入你的選擇（0或1）：";
 cin >> i;
 switch (i)
 {
 case 0:
 cre();
 exh();
 break;
 case 1:
 select = 0;
 break;
 default:
 cout << "請在0和1之間選擇！" << endl;
 }
 }
 return 0;
}