HDU3652：B-number（数位dp + 同余）

最新推荐文章于 2022-04-05 09:12:07 发布

原创最新推荐文章于 2022-04-05 09:12:07 发布 · 286 阅读

0 ·

CC 4.0 BY-SA版权

数位dp 专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一种求解特定范围内包含子串'13'且能被13整除的非负整数个数的方法。通过动态规划实现，讨论了递归搜索策略，并给出了一段C语言代码实例。

B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5639 Accepted Submission(s): 3248

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

Sample Output

Author

wqb0039

Source

2010 Asia Regional Chengdu Site —— Online Contest

题意i：求1~n中包含13且能被13整除的数的个数。

思路：见代码。

# include <stdio.h>
# include <string.h>
int a[11], dp[11][14][2][2];//dp[数位][上一位取模结果][是否已包含13][上一位是否为1]。
int dfs(int pos, int num, int pre, int sta, bool limit)
{
    if(pos==-1)
        return sta && num==0;
    if(!limit && (dp[pos][num][sta][pre==1] != -1))
        return dp[pos][num][sta][pre==1];
    int up = limit?a[pos]:9;
    int tmp = 0;
    for(int i=0; i<=up; ++i)
            tmp += dfs(pos-1, (num*10+i)%13, i, (pre==1&&i==3)||sta, limit&&(i==a[pos]));
    if(!limit)
        dp[pos][num][sta][pre==1] = tmp;
    return tmp;

}
int solve(int num)
{
    int cnt = 0;
    while(num)
    {
        a[cnt++] = num%10;
        num /= 10;
    }
    return dfs(cnt-1, 0, 0, 0, true);
}
int main()
{
    int n;
    memset(dp, -1, sizeof(dp));
    while(~scanf("%d",&n))
        printf("%d\n",solve(n));
    return 0;
}