CSU1600:Twenty-four point(DFS)

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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 91 1 1 1 5 5 5 1

Sample Output

YesNoYes

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; }

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