UVA - 10344 23 out of 5

Problem I

23 Out of 5

Input: standard input

Output: standard output

Time Limit: 1 second

Memory Limit: 32 MB

Your task is to write a program that can decide whether you can find an arithmetic expression consisting of five given numbers (1<=i<=5) that will yield the value 23.
For this problem we will only consider arithmetic expressions of the following from:

where : {1,2,3,4,5} -> {1,2,3,4,5} is a bijective function 

and  {+,-,*} (1<=i<=4)     

Input

The Input consists of 5-Tupels of positive Integers, each between 1 and 50.
Input is terminated by a line containing five zero's. This line should not be processed.

Output

For each 5-Tupel print "Possible" (without quotes) if their exists an arithmetic expression (as described above) that yields 23. Otherwise print "Impossible".

Sample Input

1 1 1 1 1 

1 2 3 4 5 

2 3 5 7 11 

0 0 0 0 0 

Sample Output

Impossible 

Possible 

Possible 

题意：

给出的数字，利用加减乘的规则看能不能变成23；

同样的利用回溯来做；

#include
#include
int a[10],vis[10];
int dfs(int cur,int tot)
{
    int i,j,t;
    if(cur==5)
    {
        if(tot==23)
            return 1;
        else
            return 0;
    }
    for(i=0;i<3;i++)
        for(j=0;j<5;j++)
            if(!vis[j])
            {
                vis[j]=1;
                if(i==0)
                    t=tot+a[j];
                else if(i==1)
                    t=tot-a[j];
                else
                    t=tot*a[j];
                if(dfs(cur+1,t))
                    return 1;
                vis[j]=0;
            }
    return 0;
}
int main()
{
    int i,j,k,ok;
    while(1)
    {
        for(i=0;i<5;i++)
            scanf("%d",&a[i]);
        if(a[0]==0)
            break;
        memset(vis,0,sizeof(vis));
        ok=0;
        for(i=0;i<5;i++)
        {
            vis[i]=1;
            if(dfs(1,a[i]))
            {
                ok=1;
                break;
            }
            vis[i]=0;
        }
        if(ok)
            printf("Possible\n");
        else
            printf("Impossible\n");
    }
    return 0;    
}

#include 
#include 
#include 
#define N 5
using namespace std;
int a[N];
int vis[N];
int flag;
void dfs(int n,int sum) {

	if (n == 5 && sum == 23) {
		flag = 1;	
		return ;
	}
	if (flag) return ; // 没有这一步会超时
	else {
			for (int j = 0; j < N; j++) {
				if (!vis[j]) { 
					vis[j] = 1; 
					if (n == 0) dfs(n+1,a[j]);
					else {
						dfs(n+1,sum + a[j]);
						dfs(n+1,sum - a[j]);
						dfs(n+1,sum * a[j]);
					}
					vis[j] = 0;
				}	
			}
	}
	
}

int main() {
	while (1){
		for (int i = 0; i < N; i++) 
			scanf("%d",&a[i]);
		if (a[0] == 0 && a[1] == 0 && a[2] == 0 && a[3] == 0 && a[4] == 0)
			break;
		memset(vis,0,sizeof(vis));
		flag = 0;
		dfs(0,0);
		if (flag) printf("Possible\n");
		else  printf("Impossible\n");
	}

}