http://acm.hdu.edu.cn/showproblem.php?pid=3652

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

/***
hdu 3652  数位dp（含13且被能被13整除的数）
题目大意：求出给定区间内的数字含有“13”并且能被13整除的个数
解题思路：记忆化搜索。
           dp[i][j][k][z]：i:处理的数位，j:该数对13取模以后的值，k:是否已经包含13,z结尾的数
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

int dp[10][15][2][10],bit[10];

int dfs(int pos,int mod,int t,int now,int flag)
{
    if(pos==-1)return mod==0&&t;
    if(!flag&&dp[pos][mod][t][now]!=-1)return dp[pos][mod][t][now];
    int end=flag?bit[pos]:9;
    int ans=0;
    for(int i=0;i<=end;i++)
    {
        ans+=dfs(pos-1,(mod*10+i)%13,t||(now==1&&i==3),i,flag&&(i==end));
    }
    if(!flag)dp[pos][mod][t][now]=ans;
    return ans;
}

int solve(int n)
{
    int len=0;
    while(n)
    {
        bit[len++]=n%10;
        n/=10;
    }
    return dfs(len-1,0,0,0,1);
}

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