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" title="24点游戏">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<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

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