hdu 2089 数位dp

数位dp  #1

#include <cstdio>#include <cstring>#include <iostream>#include <vector>#include <algorithm>using namespace std ;typedef long long ll ;int a[20] ;int dp[20][2] ;int dfs( int pos , int pre ,int sta , bool limit){    if( pos <= -1) return 1 ;    if( !limit && dp[pos][sta] != -1 ) return dp[pos][sta] ;    int up = limit ? a[pos] : 9 ;    int temp = 0 ;    for( int i = 0 ; i <= up ; i ++ ){        if( pre == 6 && i == 2) continue ;        if( i == 4 ) continue ;        temp += dfs(pos-1 , i , i==6 , limit && i==a[pos]);    }    if( !limit ) dp[pos][sta] = temp ;    return temp ;}int solve( int x ){    int pos = 0 ;    while( x ){        a[pos ++] = x % 10 ;        x /= 10 ;    }    int ans = dfs(pos - 1 , -1 , 0 , true ) ;    return ans ;}int main(){    int n , m   ;    while( ~ scanf("%d %d",&n , &m)  ){        memset( dp , -1 , sizeof( dp )) ;        if( !n && !m) break ;        //for( int i = 0 ; i <= N ; i ++ ) G[i].clear() ;        //memset( centroid , 0 , sizeof( centroid )) , memset(subtree_size , 0 , sizeof(subtree_size)) ;        printf("%d\n" , solve(m) - solve(n-1));    }    return 0 ;}