uva10344

就是枚举+ 全排列

一开始写复杂了，只要每次传进去写三次+-* 就行了

题目：

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

代码：

#include <iostream>
#include <algorithm>
using namespace std;

char opera[]={'*','+','-'};
int arr[6];
bool flag=false;
void dfs(int v,int cur,int u)
{
    if(cur>5)
    {
        if(v==23 && !flag)
        {
            flag=true;
            cout<<"Possible"<<endl;
        }
        return ;
    }

   dfs(v+arr[u],cur+1,u+1);
   dfs(v-arr[u],cur+1,u+1);
   dfs(v*arr[u],cur+1,u+1);

}


int main()
{
    while(1)
    {
        for(int i=0;i<5;i++)
        {
            cin>>arr[i];
        }
        if(arr[0]==0 && arr[1]==0&&arr[2]==0&&arr[3]==0&&arr[4]==0)
        {
            break;
        }
        sort(arr,arr+5);
        do
        {
                dfs(arr[0],1,1);

                if(flag)break;
        }while(next_permutation(arr,arr+5));
        if(!flag )
            cout<<"Impossible"<<endl;

        flag=false;
    }

    return 0;
}