csu 1600: Twenty-four point

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1600: Twenty-four point

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 490  Solved: 78
[Submit][Status][Web Board]

Description

Given four numbers, can you get twenty-four through the addition, subtraction, multiplication, and division? Each number can be used only once.

Input

The input consists of multiple test cases. Each test case contains 4 integers A, B, C, D in a single line (1 <= A, B, C, D <= 13).

Output

For each case, print the “Yes” or “No”. If twenty-four point can be get, print “Yes”, otherwise, print “No”.

Sample Input

2 2 3 9
1 1 1 1 
5 5 5 1

Sample Output

Yes
No
Yes

HINT

For the first sample, (2/3+2)*9=24.

Source

 

思路：

dfs，把数组当参数，4个数先选2个算一下，然后变成三个数，传入当下一层，避免了麻烦的选数

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <stack>
#include <cctype>
#include <vector>
#include <cmath>
#include <map>
#include <queue>

#define ll long long
#define eps 1e-8

using namespace std;

double a[5];

int dfs(double p[],int len)
{
    if(len==1){
        if(fabs(p[0]-24)<eps){
            return 1;
        }
        else{
            return 0;
        }
    }
    double f[5];
    int cou;
    int i,j,k;
    for(i = 0;i <len-1;i++){
        for(j=i+1;j<len;j++){
            cou=0;
            for(k=0;k<len;k++){
                if(k!=i && k!=j){
                    f[cou]=p[k];cou++;
                }
            }
            f[cou]=p[i]+p[j];
            if(dfs(f,cou+1)) return 1;
            f[cou]=p[i]-p[j];
            if(dfs(f,cou+1)) return 1;
            f[cou]=p[j]-p[i];
            if(dfs(f,cou+1)) return 1;
            f[cou]=p[i]*p[j];
            if(dfs(f,cou+1)) return 1;
            if(p[j]!=0){
                f[cou]=p[i]/p[j];
                if(dfs(f,cou+1)) return 1;
            }
            if(p[i]!=0){
                f[cou]=p[j]/p[i];
                if(dfs(f,cou+1)) return 1;
            }
        }
    }
    return 0;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //scanf("%d",&T);
    //for(int ccnt=1;ccnt<=T;ccnt++){
    while(scanf("%lf%lf%lf%lf",&a[0],&a[1],&a[2],&a[3])!=EOF){
        if(dfs(a,4) == 1){
            printf("Yes
");
        }
        else{
            printf("No
");
        }
    }
    return 0;
}