HDU3652 B-number 数位DP

A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string "13" and can be divided by 13. For example, 130 and 2613 are wqb-numbers, but 143 and 2639 are not. Your task is to calculate how many wqb-numbers from 1 to n for a given integer n.

Input

Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).

Output

Print each answer in a single line.

Sample Input

131002001000

Sample Output

1122

题意:求既能被13整除，并且有13字串的数字。

思路：将mark表示状态，mark == 1 表示 无13且有1，  mark == 2 表示有13，mark == 0 表示无13且没有1。然后再用sta表示是否能够被13整除即可。

Code:

#include <iostream>#include <cstdio>#include <string.h>#define LL long longusing namespace std;const int MOD = 1e9+7;int a[25];LL dp[25][25][3];LL dfs( int pos , int sta , int mark , bool limit ){if( pos == -1 ) return mark == 2 && sta == 0 ;if( !limit && dp[pos][sta][mark] != -1 ) return dp[pos][sta][mark];LL ans = 0 ;int up = limit ? a[pos] : 9 ;for( int i = 0 ; i <= up ; i ++ ){int temp = ( sta * 10 + i ) % 13;int tmp = mark;if( mark == 0 && i == 1 ) tmp = 1;if( mark == 1 && i != 1 ) tmp = 0;if( mark == 1 && i == 3 ) tmp = 2; ans += dfs( pos - 1 , temp , tmp , limit && i == up ) ;}if( !limit )  dp[pos][sta][mark] = ans ;return ans ;}LL solve( LL x ){int tot = 0 ;while( x ){a[tot++] = x % 10 ;x /= 10;}return dfs( tot - 1 , 0 , 0 , true );}int main(){//ios_base::sync_with_stdio(false); cin.tie(0);memset(dp,-1,sizeof(dp));LL n;while(~scanf("%lld",&n))printf("%lld\n",solve(n));return 0;}