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[];
 
 int dfs(double p[],int len)
 {
     if(len==){
         if(fabs(p[]-)<eps){
             return ;
         }
         else{
             return ;
         }
     }
     double f[];
     int cou;
     int i,j,k;
     for(i = ;i <len-;i++){
         for(j=i+;j<len;j++){
             cou=;
             for(k=;k<len;k++){
                 if(k!=i && k!=j){
                     f[cou]=p[k];cou++;
                 }
             }
             f[cou]=p[i]+p[j];
             if(dfs(f,cou+)) return ;
             f[cou]=p[i]-p[j];
             if(dfs(f,cou+)) return ;
             f[cou]=p[j]-p[i];
             if(dfs(f,cou+)) return ;
             f[cou]=p[i]*p[j];
             if(dfs(f,cou+)) return ;
             if(p[j]!=){
                 f[cou]=p[i]/p[j];
                 if(dfs(f,cou+)) return ;
             }
             if(p[i]!=){
                 f[cou]=p[j]/p[i];
                 if(dfs(f,cou+)) return ;
             }
         }
     }
     return ;
 }
 
 int main()
 {
     //freopen("in.txt","r",stdin);
     //scanf("%d",&T);
     //for(int ccnt=1;ccnt<=T;ccnt++){
     while(scanf("%lf%lf%lf%lf",&a[],&a[],&a[],&a[])!=EOF){
         if(dfs(a,) == ){
             printf("Yes\n");
         }
         else{
             printf("No\n");
         }
     }
     return ;
 }