题目43：24 Point game

最新推荐文章于 2022-11-12 22:35:17 发布

杨林峰的笔记

最新推荐文章于 2022-11-12 22:35:17 发布

阅读量295

点赞数

CC 4.0 BY-SA版权

分类专栏：深度搜索 C++ ACM 算法文章标签：深度搜索

本文链接：https://blog.youkuaiyun.com/qq_25245961/article/details/78181854

算法同时被 3 个专栏收录

74 篇文章

订阅专栏

C++

70 篇文章

订阅专栏

ACM

39 篇文章

订阅专栏

本文介绍了一款经典游戏——24点游戏的求解算法。通过深度优先搜索模拟所有可能的运算组合来判断给定的几个数字是否能通过加、减、乘、除运算得到目标数值24。文章提供了完整的C语言实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：

http://acm.nyist.net/JudgeOnline/problem.php?pid=43

描述

There is a game which is called 24 Point game.
In this game , you will be given some numbers. Your task is to find an expression which have all the given numbers and the value of the expression should be 24 .The expression mustn’t have any other operator except plus,minus,multiply,divide and the brackets.
e.g. If the numbers you are given is “3 3 8 8”, you can give “8/(3-8/3)” as an answer. All the numbers should be used and the bracktes can be nested.
Your task in this problem is only to judge whether the given numbers can be used to find a expression whose value is the given number。

输入

The input has multicases and each case contains one line
The first line of the input is an non-negative integer C(C<=100),which indicates the number of the cases.
Each line has some integers,the first integer M(0<=M<=5) is the total number of the given numbers to consist the expression,the second integers N(0<=N<=100) is the number which the value of the expression should be.
Then,the followed M integer is the given numbers. All the given numbers is non-negative and less than 100

输出

For each test-cases,output “Yes” if there is an expression which fit all the demands,otherwise output “No” instead.

样例输入

2
4 24 3 3 8 8
3 24 8 3 3

样例输出

Yes
No

算法思想：

这是一道经典的深度搜索的题目，使用深度搜索来模拟所有的括号和加减乘除。

源代码


#include<stdio.h>
#include<string.h>
#include<math.h>
int n, vis[10];
int flag;
double a[10], aim;

void dfs(double num);


int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        flag = 0;
        scanf("%d%lf", &n, &aim);
        for (int i = 0; i<n; i++)
            scanf("%lf", &a[i]);
        for (int i = 0; i<n; i++)
        {
            memset(vis, 0, sizeof(vis));
            vis[i] = 1;
            dfs(a[i]);
            if (flag)
            {
                break;
            }
        }
        if (flag)
            printf("Yes\n");
        else
            printf("No\n");
    }

    return 0;
}


void dfs(double sum)
{
    int cnt = 0;
    for (int i = 0; i<n; i++)
        if (vis[i])
            cnt++;
    if (cnt == n)
    {
        if (fabs(aim - sum)<0.0000001)
            flag = 1;
        return;
    }
    for (int i = 0; i<n; i++)
    {
        if (!vis[i] && !flag)
        {
            vis[i] = 1;
            dfs(a[i] + sum);
            dfs(a[i] - sum);
            dfs(a[i] * sum);
            dfs(sum - a[i]);
            if (a[i] != 0.0)
                dfs(sum / a[i]);
            if (sum != 0.0)
                dfs(a[i] / sum);
            vis[i] = 0;
        }
    }
}