【BZOJ】1026: [SCOI2009]windy数 数位DP

传送门：【BZOJ】1026: [SCOI2009]windy数

题目分析：数位DP水题。

代码如下：

#include <stdio.h>
#include <cstring>
#include <algorithm>

#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 clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 11 ;

int num[MAXN] ;
int dp[MAXN][MAXN][2][2] ;
int vis[MAXN][MAXN][2][2] ;
int a , b , n ;

int dfs ( int cur , int j , int flag , int num[] , int pre_zero ) {
	if ( cur == 0 ) return 1 ;
	if ( vis[cur][j][flag][pre_zero] ) return dp[cur][j][flag][pre_zero] ;
	vis[cur][j][flag][pre_zero] = 1 ;
	dp[cur][j][flag][pre_zero] = 0 ;
	if ( flag ) {
		rep ( i , 0 , 10 ) {
			if ( pre_zero ) {
				dp[cur][j][flag][pre_zero] += dfs ( cur - 1 , i , 1 , num , !i ) ;
			} else if ( abs ( j - i ) >= 2 ) {
				dp[cur][j][flag][pre_zero] += dfs ( cur - 1 , i , 1 , num , 0 ) ;
			}
		}
	} else {
		rep ( i , 0 , num[cur] ) {
			if ( pre_zero ) {
				dp[cur][j][flag][pre_zero] += dfs ( cur - 1 , i , 1 , num , !i ) ;
			} else if ( abs ( j - i ) >= 2 ) {
				dp[cur][j][flag][pre_zero] += dfs ( cur - 1 , i , 1 , num , 0 ) ;
			}
		}
		if ( pre_zero ) {
			dp[cur][j][flag][pre_zero] += dfs ( cur - 1 , num[cur] , 0 , num , !num[cur] ) ;
		} else if ( abs ( j - num[cur] ) >= 2 ) {
			dp[cur][j][flag][pre_zero] += dfs ( cur - 1 , num[cur] , 0 , num , 0 ) ;
		}
	}
	return dp[cur][j][flag][pre_zero] ;
}

int solve ( int number ) {
	clr ( vis , 0 ) ;
	n = 0 ;
	while ( number ) {
		num[++ n] = number % 10 ;
		number /= 10 ;
	}
	int ans = dfs ( n , 0 , 0 , num , 1 ) ;
	//printf ( "%d\n" , ans ) ;
	return ans ;
}

int main () {
	while ( ~scanf ( "%d%d" , &a , &b ) ) printf ( "%d\n" , solve ( b ) - solve ( a - 1 ) ) ;
	return 0 ;
}