hdu 4734 数位dp #2

can you hear me ... 数位dp 。。 

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>using namespace std;typedef  long long  ll ;int f( int x ){    if( x == 0 ) return 0 ;    int ans = f( x / 10 );    return ans * 2 + x % 10;}int a[20] , tempsum ;int dp[20][20000] ;int dfs( int pos , int sum , bool limit){    if( pos == -1 ) {return sum <= tempsum ;}    if( sum > tempsum ) return 0 ;    if( !limit && dp[pos][tempsum - sum ] != -1) return dp[pos][tempsum -  sum ] ;    int tmp = 0 ;    int up = limit ? a[pos] : 9 ;    for( int i = 0 ; i <= up ; i ++ ){        tmp += dfs( pos - 1 , sum + i * (1 << pos) , limit && a[pos] == i );    }    if( !limit ) dp[pos][tempsum - sum ] = tmp ;    return tmp ;}int solve(int x){    int tmp = 0 ;    while( x ){        a[tmp ++ ] = x % 10 ;        x /= 10 ;    }    return dfs( tmp - 1 , 0 , true  ) ;}int main(){    int  n , m , kase,t = 1;    scanf("%d",&kase);    memset( dp , -1 , sizeof( dp ));    while( kase -- )    {        scanf("%d %d" , & n , &m) ;        tempsum = f(n) ;        printf("Case #%d: %d\n",t ++ , solve(m) );    }return 0;}