usaco1.34Prime Cryptarithm

之前做的一道题，忘记发了，纠结了好久，把题意想复杂了，就是直接枚举*号位置上的数。

/*
 ID: your_id_here
 PROG: crypt1
 LANG: C++
 */
#include <iostream>
#include<cstdio>
#include<string.h>
using namespace std;
int count,f[100];
int judge(int s)
{
    int i,flag = 1,m = s;
    while(s)
    {
        i = s%10;
        s = s/10;
        if(!f[i])
        {
            flag = 0;
            break;
        }
    }
    if(flag)
    return 1;
    return 0;
}
int main()
{
    freopen("crypt1.in","r",stdin);
    freopen("crypt1.out","w",stdout);
    int n,i,j,k,g,h,num[15];
    cin>>n;
    count = 0;
    for(i = 1 ; i <= n ; i++)
    {
        cin>>num[i];
        f[num[i]] = 1;
    }
    for(i = 1; i <= n ; i++)
    for(j = 1;j <= n; j++)
    for(k = 1; k <= n ; k++)
    {
        int x = num[i]+num[j]*10+num[k]*100;
        for(g = 1 ; g <= n ; g++)
        for(h = 1; h <= n ; h++)
        {
            int s1 = x*num[g];
            int s2 = x*num[h];
            int y = num[g]+num[h]*10;
            int s = x*y;
            if(judge(s1)&&judge(s2)&&judge(s))
            {
                if(s<10000&&s>=1000&&s1>=100&&s1<1000&&s2>=100&&s2<1000)
                    count++;
            }
        }
    }
    cout<<count<<endl;
    fclose(stdin);
    fclose(stdout);
    return 0;
}