学步园

暴力枚举就可以，但是暴力枚举也是有小技巧的，比如，先判断再枚举，基本上可以省下一层循环的浪费

code

/*
ID: yueqiq
PROG: crypt1
LANG: C++
*/
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("crypt1.in","r",stdin)
#define writef freopen("crypt1.out","w",stdout)
const int maxn = 20005;
const int INF = 0x3fffffff;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
const double pi = acos(-1.0);
using namespace std;
int N,a[9],cnt;
bool vis[10];
bool check(int s){
    while(s){
        int t=s%10;
        s/=10;
        if(!vis[t]) return false;
    }
    return true;
}
int main(){
    readf;
    writef;
    SD(N);
    FF(i,N){
        SD(a[i]);
        vis[a[i]]=true;
    }
    FF(i,N) FF(j,N) FF(k,N) FF(l,N){
        int top=(a[i]*100+a[j]*10+a[k])*a[l];
        if(top>999 || !check(top)) continue;
        FF(m,N){ //将这个循环写进来是一个不小的优化
            int buttom=(a[i]*100+a[j]*10+a[k])*a[m];
            if(buttom>999 || !check(buttom)) continue;
            int ans=top+buttom*10;
            if(ans>9999 || !check(ans)) continue;
            cnt++;
        }
    }
    PD(cnt);LN;
    return 0;
}