hdu 3652 B-number (数位DP)

题目地址：点击打开链接

题意：求1-n中包含13且能被13整除的数的数量。

思路：dp[i][j][k]，i表示位数，j表示余数，k表示末尾是1、末尾不是1、含有13.

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 15;
int a[maxn], dp[maxn][maxn][3];

int dfs(int pos, int pre, int mod, int limit)
{
    if(pos == -1) return (pre==2 && !mod);
    if(!limit && dp[pos][mod][pre] != -1) return dp[pos][mod][pre];
    int up = limit ? a[pos] : 9;
    int tmp = 0;
    for(int i = 0; i <= up; i++)
    {
        int tmod = (mod*10+i)%13;
        if((pre == 1 && i == 3) || pre == 2)
            tmp += dfs(pos-1, 2, tmod, limit&&a[pos]==i);
        else
            tmp += dfs(pos-1, i==1, tmod, limit&&a[pos]==i);
    }
    if(!limit) dp[pos][mod][pre] = tmp;
    return tmp;
}

int solve(int x)
{
    int pos = 0;
    while(x)
    {
        a[pos++] = x%10;
        x /= 10;
    }
    return dfs(pos-1, 0, 0, 1);
}

int main(void)
{
    int n;
    memset(dp, -1, sizeof(dp));
    while(cin >> n)
        printf("%d\n", solve(n));
    return 0;
}

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5789    Accepted Submission(s): 3332

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