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

 

题意：求小于等于n的可以被13整除的并且包含13的数有多少个

用容斥原理n减去不能被13整除的减去不含13的，中间多减了，需要加上既不包含13也不能被13整除的数的个数就是答案

代码如下：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
ll dp[50][50][50];
ll dp1[25][50];
int a[25];
ll dfs(int pos,int pre,int sta,int sum,int limit)
{
    if(pos<0)return sum?1:0;
    if(!limit&&dp[pos][sum][sta]!=-1)
        return dp[pos][sum][sta];
    int up=limit?a[pos]:9;
    ll ans=0;
    for(int i=0;i<=up;i++)
    {
        if(pre==1&&i==3)continue;
        ans+=dfs(pos-1,i,i==1,(sum*10+i)%13,limit&&i==up);
    }
    if(!limit) dp[pos][sum][sta]=ans;
    return ans;
}
ll dfs1(int pos,int pre,int sta,int limit)
{
    if(pos<0)return 1;
    if(!limit&&dp1[pos][sta]!=-1)
        return dp1[pos][sta];
    int up=limit?a[pos]:9;
    ll ans=0;
    for(int i=0;i<=up;i++)
    {
        if(pre==1&&i==3)continue;
        ans+=dfs1(pos-1,i,i==1,limit&&(i==up));
    }
    if(!limit) dp1[pos][sta]=ans;
    return ans;
}
ll solve(int n)
{
    int t=0;
    ll tt=n;
    ll ans3=n-n/13; //不能整除13的
    while(n)
    {
        a[t++]=n%10;
        n/=10;
    }
    ll ans1=dfs(t-1,-1,0,0,1); //既不能被13整除，也不包含13的
    ll ans2=dfs1(t-1,-1,0,1)-1; //不含13的
    return tt-ans2-ans3+ans1;//容斥
}
int main()
{
    int n;
    memset(dp,-1,sizeof(dp));
    memset(dp1,-1,sizeof(dp1));
    while(~scanf("%d",&n))  printf("%lld\n",solve(n)); 


    return 0;
}