学步园

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

求包含13且是13的倍数的个数
状态很好设计
dp[i][rest][k][e] 表示i位数,最高为为e,模13为k,是否包含13
然后套上漂亮的数位dp板子
算是熟悉一下这个写法了

/*************************************************************************
    > File Name: hdu3652.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月22日 星期日 14时09分09秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int dp[11][14][2][11];
int bit[12];

int dfs (int cur, int rest, bool t, int e, bool flag)
{
    if (cur == -1)
    {
        return t && (!rest);
    }
    if (~dp[cur][rest][t][e] && !flag)
    {
        return dp[cur][rest][t][e];
    }
    int end = flag ? bit[cur] : 9;
    int ans = 0;
    for (int i = 0; i <= end; ++i)
    {
        ans += dfs (cur - 1, (rest * 10 + i) % 13, t || (e == 1 && i == 3), i, flag && (i == end));
    }
    if (!flag)
    {
        dp[cur][rest][t][e] = ans;
    }
    return ans;
}

int calc (int n)
{
    int cnt = 0;
    while (n)
    {
        bit[cnt++] = n % 10;
        n /= 10;
    }
    return dfs (cnt - 1, 0, 0, 0, 1);
}

int main ()
{
    int n;
    memset (dp, -1, sizeof(dp));
    while (~scanf("%d", &n))
    {
        printf("%d\n", calc (n));
    }
    return 0;
}