CSU1600:Twenty-four point(DFS)

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

题意:四个数字，通过四则运算看能不能得到24点

思路:对于如此小量的数，直接暴搜，你不需要怀疑人生

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,sizeof(a))
#define w(a) while(a)
#define LL long long
const double pi = acos(-1.0);
#define Len 200005
#define mod 19999997
const int INF = 0x3f3f3f3f;
#define exp 1e-8 

double a[5]; 

bool dfs(double *a,int n)
{
    if(n==1) return fabs(a[0]-24)<exp;
    int i,j,k,l;
    double tem[5];
    up(i,0,n-1)
    {
        up(j,i,n-1)
        {
            if(i==j) continue;
            int len = 0;
            up(l,0,n-1)
            {
                if(l!=i && l!=j)
                    tem[len++]=a[l];
            }
            tem[len] = a[i]+a[j];
            if(dfs(tem,n-1)) return true;
            tem[len] = a[i]-a[j];
            if(dfs(tem,n-1)) return true;
            tem[len] = a[j]-a[i];
            if(dfs(tem,n-1)) return true;
            tem[len] = a[i]*a[j];
            if(dfs(tem,n-1)) return true;
            if(a[i]!=0)
            {
                tem[len] = a[j]/a[i];
                if(dfs(tem,n-1)) return true;
            }
            if(a[j]!=0)
            {
                tem[len] = a[i]/a[j];
                if(dfs(tem,n-1)) return true;
            }
        }
    }
    return false;
} 

int main()
{
    w(~scanf("%lf%lf%lf%lf",&a[0],&a[1],&a[2],&a[3]))
    {
        if(dfs(a,4))
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}