Lightoj 1021 数位DP

dp[statu][mod] 表示还剩的位为statu且之前余数为mod时的个数

简单的数位DP

AC代码如下：

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

long long dp[( 1 << 16 ) + 10][22];
int number[20], cnt;
int base, K;

long long DFS( int statu, int mod, int num ){
	if( num == cnt ){
		return mod == 0;
	}
	if( dp[statu][mod] != -1 ){
		return dp[statu][mod];
	}
	long long ans = 0;
	for( int i = 0; i < cnt; i++ ){
		if( statu & ( 1 << i ) ){
			ans += DFS( statu ^ ( 1 << i ), ( mod * base + number[i] ) % K, num + 1 );
		}
	}
	return dp[statu][mod] = ans;
}

int main(){
	int T, Case = 1;

	scanf( "%d", &T );
	while( T-- ){
		scanf( "%d%d", &base, &K );
		char s[20];
		scanf( "%s", s );
		cnt = 0;
		for( int i = 0; s[i]; i++ ){
			if( s[i] >= '0' && s[i] <= '9' ){
				number[cnt++] = s[i] - '0';
			}else if( s[i] >= 'A' && s[i] <= 'F' ){
				number[cnt++] = s[i] - 'A' + 10;
			}
		}
		memset( dp, -1, sizeof( dp ) );
		cout << "Case " << Case++ << ": " << DFS( ( 1 << cnt ) - 1, 0, 0 ) << endl;
	}
	return 0;
}