uva10344

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：输入5个整数，按照某种顺序排列后依次进行+, -或者*，使得最终结果为23。判断是否有解
// 算法：回溯

time 1.692

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int a[5];
int b[5];
char op[5];
int vis[5];
int possible;


void dfs(int d, int s)
{
    if(d==5)
    {
        if(s==23)
        {
            possible=1;
#ifndef ONLINE_JUDGE
            printf("(((%d ", b[0]);
            for(int i=1;i<4;i++)
                printf("%c %d) ", op[i], b[i]);
            printf("%c %d ", op[4], b[4]);
            printf("
");
#endif
        }

        return;
    }

    for(int i=0;i<5;i++)
    {
        //第一个数
        if(d==0)
        {
            if(!vis[i])
            {
                vis[i]=1;
                b[d]=a[i];
                dfs(d+1, a[i]);
                vis[i]=0;
            }
        }
        else
        {
            if(!vis[i])
            {
                vis[i]=1;
                b[d]=a[i];
                op[d]='+';
                dfs(d+1, s+a[i]);
                op[d]='-';
                dfs(d+1, s-a[i]);
                op[d]='*';
                dfs(d+1, s*a[i]);
                vis[i]=0;
            }
        }
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("./uva10344.in", "r", stdin);
#endif
    while(scanf("%d%d%d%d%d", a, a+1, a+2, a+3, a+4)==5 
            && (a[0] || a[1] || a[2] || a[3] || a[4]))
    {
        possible=0;
        dfs(0, 0);
        if(possible)
            puts("Possible");
        else
            puts("Impossible");
    }

    return 0;
}

找到后立刻返回，time 0.945

学习点，dfs时判断扩展点的返回值，如果已经成功，就直接返回。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int a[5];
int vis[5];


bool dfs(int d, int s)
{
    if(d==5)
        return s==23;

    for(int i=0;i<5;i++) if(!vis[i])
    {

        vis[i]=1;
        //第一个数
        if(d==0) { if(dfs(d+1, a[i])) return true; }
        else
        {
            if(dfs(d+1, s+a[i])) return true;
            if(dfs(d+1, s-a[i])) return true;
            if(dfs(d+1, s*a[i])) return true;
        }
        vis[i]=0;
    }
    return false;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("./uva10344.in", "r", stdin);
#endif
    while(scanf("%d%d%d%d%d", a, a+1, a+2, a+3, a+4)==5 
            && (a[0] || a[1] || a[2] || a[3] || a[4]))
    {
        memset(vis, 0, sizeof(vis));
        if(dfs(0, 0))
            puts("Possible");
        else
            puts("Impossible");
    }

    return 0;
}

 

next_permutation 计算差点超时： 2.388

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int a[5];

bool check(int k)
{
    int s=a[0];
    for(int j=1;j<5;j++)
    {
        switch(k%3)
        {
            case 0:
                s+=a[j]; break;
            case 1:
                s-=a[j]; break;
            case 2:
                s*=a[j]; break;
        }
        k/=3;
    }
    return s==23;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("./uva10344.in", "r", stdin);
#endif
    while(scanf("%d%d%d%d%d", a, a+1, a+2, a+3, a+4)==5 
            && (a[0] || a[1] || a[2] || a[3] || a[4]))
    {
        sort(a, a+5);
        int ok=0;
        do
        {
            for(int i=0;i<81;i++) if(check(i))
            {
                ok=1;
                break;
            }
        }while(!ok && next_permutation(a, a+5));
        if(ok)
            puts("Possible");
        else
            puts("Impossible");
    }

    return 0;
}