B-number (the number of bits dp)

topic

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

Explanation

dp [pos] [sta] [ pre] [sum] represents the position pos, sta 1 130 representatives described previously in the other hand, on behalf of a former pre whether 1, sum values of the 13 preceding the representative results taken remaining.
Because the current value of the sum determines how many meet later remainder is the number 0, so the sum to add this dimension, sta and pre synthesis can also be a dimension of judgment.

#include <cstdio>
#include <cstring>
#define ll long long 
using namespace std;
ll dp[20][2][2][14];
int dig[20];
ll dfs(int pos, int sta, int pre, int sum, int limit){
	if(pos == -1) return sta && !sum ? 1:0;
	if(!limit && dp[pos][sta][pre == 1][sum] != -1) return dp[pos][sta][pre == 1][sum];
	int up = limit? dig[pos]:9;
	ll ans = 0;
	for(int i = 0; i <= up; i++){
		ans += dfs(pos-1, sta||(pre == 1 && i == 3), i, (sum*10+i)%13, limit && i == up);
	}
	if(!limit) dp[pos][sta][pre == 1][sum] = ans;
	return ans; 
}
ll slove(ll x){
	int pos = 0;
	while(x){
		dig[pos++] = x%10;
		x /= 10; 
	}
	ll ans = dfs(pos-1, 0, 0,0, 1);
	return ans;
}
int main(){
	int n;
	memset(dp, -1, sizeof(dp));
	while(~scanf("%d", &n)){
		printf("%I64d\n", slove(n));
	}
	return 0;
}