HDU - 3652 B-number

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

思路：

数位dp。dp[i][j][k]中i表示第i位，j表示第i位之前的数模13的值，k=0表示第i+1位不为1，k=1表示第i+1位为1，k=2表示第i位之前已经发现过13。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int n, num[20], dp[20][13][3];

int dfs(int pos, int mod, int state, bool limit)
{
    if (pos == -1)
	{
        return mod == 0 && state == 2;
	}

    if (!limit && dp[pos][mod][state])
	{
        return dp[pos][mod][state];
	}

    int result = 0;
	int up = limit ? num[pos] : 9;
    for (int i = 0; i <= up; i++)
    {
        int temp = state;
        if (state != 2 && i != 1)
		{
            temp = 0;
		}
        if (state == 1 && i == 3)
		{
            temp = 2;
		}
        if (i == 1 && state != 2)
		{
            temp = 1;
		}
        result += dfs(pos - 1, (mod * 10 + i) % 13, temp, limit && i == up);
    }

    if (!limit)
	{
        dp[pos][mod][state] = result;
	}

    return result;
}

int main()
{
    while (~scanf("%d", &n))
    {
        memset(num, 0, sizeof(num));
        memset(dp, 0, sizeof(dp));
        int pos = 0;
        while (n)
        {
            num[pos] = n % 10;
            n /= 10;
			pos++;
        }

        printf("%d\n", dfs(pos, 0, 0, true));
    }

    return 0;
}