HDU

速算24点

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7037    Accepted Submission(s): 1857

 

Problem Description

速算24点相信绝大多数人都玩过。就是随机给你四张牌，包括A(1),2,3,4,5,6,7,8,9,10,J(11),Q(12),K(13)。要求只用'+','-','*','/'运算符以及括号改变运算顺序，使得最终运算结果为24(每个数必须且仅能用一次)。游戏很简单，但遇到无解的情况往往让人很郁闷。你的任务就是针对每一组随机产生的四张牌，判断是否有解。我们另外规定，整个计算过程中都不能出现小数。

 

Input

每组输入数据占一行，给定四张牌。

 

Output

每一组输入数据对应一行输出。如果有解则输出"Yes"，无解则输出"No"。

 

Sample Input

A 2 3 6

3 3 8 8

 

Sample Output

Yes

No

 

思路

由于没有运算顺序, 我们可以选择任意两个数进行加减乘除操作.

 

#include <bits/stdc++.h>
using namespace std;

int a[4];
bool vis[4], OK, flag;
void dfs(int cur)
{
    /* 
    cout<<cur<<'
';
    for(int i=0;i<4;++i) {
        cout<<a[i]<<' ';
    }
    cout<<'
';*/
    if (OK)
        return;
    if (cur == 3)
    {
        for (int i = 0; i < 4; i++)
            if (vis[i] && a[i] == 24)
                OK = true;
                
        return;
    }
    for (int i = 0; i < 4; i++)
        if (vis[i])
            for (int j = i + 1; j < 4; j++)
                if (vis[j])
                {
                    vis[j] = false;
                    int x = a[i], y = a[j];
                    if (y != 0 && x % y == 0)
                        a[i] = x / y, dfs(cur + 1);
                    if (x != 0 && y % x == 0)
                        a[i] = y / x, dfs(cur + 1);
                    a[i] = x - y, dfs(cur + 1);
                    a[i] = y - x, dfs(cur + 1);
                    a[i] = x + y, dfs(cur + 1);
                    a[i] = x * y, dfs(cur + 1);
                    a[i] = x;
                    vis[j] = true;
                }
}

int READ()
{
    char c[4][4];
    int res = scanf("%s%s%s%s", c[0], c[1], c[2], c[3]);
    for (int i = 0; i < 4; i++)
    {
        if (c[i][0] == 'A')
            a[i] = 1;
        else if (c[i][0] == 'J')
            a[i] = 11;
        else if (c[i][0] == 'Q')
            a[i] = 12;
        else if (c[i][0] == 'K')
            a[i] = 13;
        else if (c[i][0] == '1')
            a[i] = 10;
        else
            a[i] = c[i][0] - '0';
    }
    memset(vis, true, sizeof(vis));
    OK = false;
    return ~res;
}

int main()
{
#ifndef ONLINE_JUDGE
   // freopen("in.txt", "r", stdin);
   // freopen("out.txt", "w", stdout);
#endif
    while (READ())
    {
        /* for(int i = 0; i < 4; i++)
            cout << a[i] << " ";
        cout << endl;*/
        dfs(0);
        puts(OK ? "Yes" : "No");
    }
}