【数位DP】G - B-number HDU - 3652

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8578    Accepted Submission(s): 5088

 

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

 

Author

wqb0039

 

Source

2010 Asia Regional Chengdu Site —— Online Contest

#include <bits/stdc++.h>

using namespace std;

int ch[20];
int dp[20][2][2][20];

int dfs(int pos, bool in, bool pre, int mod, bool limit)
{
	if (pos == 0)
		return (in && mod == 0) ? 1 : 0;

	if (!limit && dp[pos][in][pre][mod])
		return dp[pos][in][pre][mod];

	int res = 0;
	int up = limit ? ch[pos] : 9;
	for (int i = 0; i <= up; i++)
	{
		/// 将当前数 % 13 传递，类似列竖式。。
		if (pre && i == 3)
			res += dfs(pos - 1, 1, 0, (mod * 10 + i) % 13, limit && i == up);
		else
			res += dfs(pos - 1, in, i == 1, (mod * 10 + i) % 13, limit && i == up);
	}
	if (!limit)
		dp[pos][in][pre][mod] = res;

	return res;
}

int solve(int n)
{
	int cnt = 0;
	while (n > 0)
	{
		ch[++cnt] = n % 10;
		n /= 10;
	}
	return dfs(cnt, 0, 0, 0, 1);
}

int main()
{
	#ifndef ONLINE_JUDGE
		freopen("C:\\in.txt", "r", stdin);
	#endif // ONLINE_JUDGE

	int n;
	while (~scanf("%d", &n))
		printf("%d\n", solve(n));
	return 0;
}