1 #include<stdio.h>
2 #include<string.h>
3 using namespace std;
4 typedef long long ll;
5
6 int a[20];
7 int dp[20][15][3];
8
9 //pos:当前位置
10 //mod：余数
11 //have ：0：前一位不是1 1：前一位是1 2：有13
12 ll dfs(int pos,int mod,int have,bool limit){
13 if(pos==-1) return mod == 0 && have == 2;
14 if(!limit && dp[pos][mod][have] != -1)
15 return dp[pos][mod][have];//没有上限并且已被访问过
16 int up = limit?a[pos]:9;
17 ll ans = 0,i;
18 for(i = 0;i <= up;i++){
19 int modx = (mod*10+i)%13;
20 int havex = have;
21 if(have == 0 && i==1) havex=1;
22 if(have == 1 && i!=1) havex=0;
23 if(have == 1 && i == 3) havex=2;
24 ans += dfs(pos-1, modx, havex, limit && i == a[pos]);
25 }
26 if(!limit) dp[pos][mod][have] = ans;
27 return ans;
28 }
29
30 ll solve(ll x){
31 int pos=0;
32 while(x){
33 a[pos++]=x%10;
34 x/=10;
35 }
36 return dfs(pos-1,0,0,true);
37 }
38 int main(){
39 ll n;
40 memset(dp,-1,sizeof(dp));
41 while(scanf("%lld",&n)!=EOF){
42 ll ans = solve(n);
43 printf("%lld\n",ans;
44 }
45 return 0;
46 }
