SDUT 3297 DFS, sdut3297dfs

SDUT 3297 DFS, sdut3297dfs

Wonderful 23 o'clock Time Limit: 1000 ms Memory limit: 65536 K any questions? Click Here ^_^ Description

The question is very simple. Five numbers are given. You can use the '+', '-', and '*' operators (the operator has no priority relationship) to make the final calculation result equal to 23, the five numbers can be changed in any order.

Enter 5 numbers in the range of [1, 50]. Output: if the final calculation result is 23, the output is Yes. If not, the output is No.

Sample Input

1 1 1 1 11 2 3 4 52 3 5 7 11

Sample output

NoYesYes

Prompt

Prompt

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<map>using namespace std;int flag=0;int vis[55];int a[6];char c[]= {'+','-','*'};int DFS(int i ,int ans,int num){    if(flag||num>5)        return 0;    if(ans==23&&num==5)    {        flag=1;        return 1;    }    for(int j=0; j<5; j++)        for(int i=0; i<3; i++)        {            if(vis[j]==0)            {                vis[j]=1;                if(i==0)                    DFS(j,ans+a[j],num+1);                else if(i==1)                    DFS(j,ans-a[j],num+1);                else                    DFS(j,ans*a[j],num+1);                vis[j]=0;            }        }    return 0;}int main(){    while(~scanf("%d",&a[0]))    {        flag=0;        for(int i=1; i<5; i++)            scanf("%d",&a[i]);        for(int i=0; i<5; i++)        {            memset(vis,0,sizeof(vis));            vis[i]=1;            DFS(i,a[i],1);            if(flag)                break;        }        if(flag==0)            printf("No\n");        else            printf("Yes\n");    }} 

Copyright Disclaimer: This article is an original article by the blogger and cannot be reproduced without the permission of the blogger.