codeforces 401D (数位DP)

思路：很明显的数位dp,设dp[i][j] 表示选取数字的状态为i，模m等于j的数的个数，那么最后的答案就是dp[(1<<n)-1][0]。状态转移方程就是,dp[i|(1<<k)][(10*j+n[j])%m]+=dp[i][k]

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
const int MAXN = 18;
const int MAXM = 101;
LL dp[1 << MAXN][MAXM], d = 1;
int main(){
    int l, m, len, c[10] = {0};
    char n[20];
    cin >> n >> m;
    l = strlen(n), len = (1 << l);
    dp[0][0] = 1;
    for(int i = 0;i < l;i ++) d *= ++c[n[i] -= '0'];
    for(int i = 0;i < len;i ++){
        for(int j = 0;j < l;j ++){
            if(i & (1 << j)) continue;
            if(i || n[j]){
                for(int k = 0;k < m;k ++)
                    dp[i|(1<<j)][(k*10+n[j])%m] += dp[i][k];
            }
        }
    }
    cout << dp[len-1][0]/d << endl;
}