HDU3652[B-number]

Problem Description

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

13
100
200
1000

Sample Output

1
1
2
2

题目大意：求出在1~n的整数中，满足含有字串“13”，以及能被13整除的数的个数

题解：数位DP

DP[pos][mod][chk][now]来表示在第pos位，与十三取膜为MOD，含有或不含字串“13”， 最后一位是now时满足题目条件的整数个数。

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int a[15], dp[15][15][3][10];int dfs( int pos, int mod, int chk, int now,int flg ){    if ( pos == -1 ) return (!mod) && chk;    if ( !flg && dp[pos][mod][chk][now] != -1 ) return dp[pos][mod][chk][now];    int up = flg ? a[pos] : 9, ret = 0;    for ( int i = 0; i <= up; i++) ret += dfs( pos-1, (mod*10+i)%13, chk || ( now == 1 && i == 3), i, flg && ( i == up) );    if ( !flg ) dp[pos][mod][chk][now] = ret;    return ret;} int solve( int x ){    int pos = 0;    for ( ; x; a[pos++] = x % 10, x /= 10);    return dfs( pos-1, 0, 0, 0, 1);}int main(){    int m;    memset( dp, -1, sizeof(dp));    while ( ~scanf( "%d", &m) ){        printf( "%d\n", solve( m ));    }    return 0;}