CSU

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 91 1 1 1 5 5 5 1

Sample Output

YesNoYes

Hint

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

解题思路：也是参考了大神的代码，因为数据较小，其实可以直接去dfs得出答案

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;double EPS = 1e-10;double num[4];bool flag;bool equel(double a, double b){    if(fabs(a - b) <= EPS)        return true;    return false;}void DFS(int n){    if(flag || (n == 1 && equel(num[0], 24)))    {        flag = true;        return;    }    for(int i = 0; i < n; i++)    {        for(int j = i + 1; j < n; j++)        {            double c1 = num[i], c2 = num[j];            num[j] = num[n - 1];            num[i] = c1 + c2;            DFS(n - 1);            num[i] = c1 - c2;            DFS(n - 1);            num[i] = c2 - c1;            DFS(n - 1);            num[i] = c1 * c2;            DFS(n - 1);            if(!equel(c2, 0))            {                num[i] = c1 / c2;                DFS(n - 1);            }            if(!equel(c1, 0))            {                num[i] = c2 / c1;                DFS(n - 1);            }            num[i] = c1;            num[j] = c2;        }    }}int main(){    while(scanf("%lf %lf %lf %lf", &num[0], &num[1], &num[2], &num[3]) != EOF)    {        flag = false;        DFS(4);        printf("%s\n", flag ? "Yes" : "No");    }}