HDU 3652 B-number（*数位DP）

G - B-number

 HDU - 3652 

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

Hint

含有数字13和能够被13整除的数的个数

#include <bits/stdc++.h>
using namespace std;

int dp[12][15][2][10];///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,判断13所用
int bit[12];

int dfs(int pos,int num,bool t,int pre,bool flag)
{
    if(pos==-1)return t&&(num==0);
    if(!flag && dp[pos][num][t][pre]!=-1)
        return dp[pos][num][t][pre];


    int end=flag?bit[pos]:9;
    int ans=0;
    for(int i=0;i<=end;i++)
        ans+=dfs(pos-1,(num*10+i)%13,t||(pre==1&&i==3),i,flag&&(i==end));


    if(!flag)dp[pos][num][t][pre]=ans;
    return ans;
}
int calc(int n)
{
    int pos=0;
    while(n)
    {
        bit[pos++]=n%10;
        n/=10;
    }
    return dfs(pos-1,0,0,0,1);
}
int main()
{
    int n;
    memset(dp,-1,sizeof(dp));
    while(scanf("%d",&n)==1)
        printf("%d\n",calc(n));
    return 0;
}