【HDU】3652 B-number 数位DP

传送门：【HDU】3652 B-number

题目分析：记录数字对13取模后的状态。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
  
typedef long long LL ;
  
#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define clr( a , x ) memset ( a , x , sizeof a )
#define cpy( a , x ) memcpy ( a , x , sizeof a )

const int MAXN = 65 ;

int dp[MAXN][10][2][2][13] ;
int ten[MAXN] ;
int vis[MAXN][10][2][2][13] , Time ;
int num[MAXN] ;
int n ;

int dfs ( int cur , int j , int flag , int has , int number ) {
	//if ( cur == -1 ) printf ( "%d\n" , number ) ;
	if ( cur == -1 ) return has && number % 13 == 0 ;
	int& tmp = dp[cur][j][flag][has][number * ten[cur] % 13] ;
	if ( vis[cur][j][flag][has][number * ten[cur] % 13] == Time ) return tmp ;
	vis[cur][j][flag][has][number * ten[cur] % 13] = Time ;
	tmp = 0 ;
	if ( flag ) {
		rep ( i , 0 , 10 ) {
			if ( j == 1 && i == 3 ) tmp += dfs ( cur - 1 , i , 1 , 1 , number * 10 + i ) ;
			else tmp += dfs ( cur - 1 , i , 1 , has , number * 10 + i ) ;
		}
	} else {
		rep ( i , 0 , num[cur] ) {
			if ( j == 1 && i == 3 ) tmp += dfs ( cur - 1 , i , 1 , 1 , number * 10 + i ) ;
			else tmp += dfs ( cur - 1 , i , 1 , has , number * 10 + i ) ;
		}
		if ( j == 1 && num[cur] == 3 ) tmp += dfs ( cur - 1 , num[cur] , 0 , 1 , number * 10 + num[cur] ) ;
		else tmp += dfs ( cur - 1 , num[cur] , 0 , has , number * 10 + num[cur] ) ;
	}
	return tmp ;
}

void solve () {
	int n1 = 0 ;
	while ( n ) {
		num[n1 ++] = n % 10 ;
		n /= 10 ;
	}
	++ Time ;
	clr ( dp , 0 ) ;
	int count = dfs ( n1 - 1 , 0 , 0 , 0 , 0 ) ;
	printf ( "%d\n" , count ) ;
}

int main () {
	clr ( vis , 0 ) ;
	Time = 0 ;
	ten[0] = 1 ;
	rep ( i , 1 , MAXN ) ten[i] = ten[i - 1] * 10 ;
	while ( ~scanf ( "%d" , &n ) ) solve () ;
	return 0 ;
}