[HDU 3652] B-number

最新推荐文章于 2019-09-29 17:40:00 发布

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最新推荐文章于 2019-09-29 17:40:00 发布

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CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/hate13/p/4115398.html

本文介绍了一种特殊类型的数字——B-number，并通过数位DP算法解决了一个相关问题：计算从1到n范围内所有B-number的数量。B-number定义为非负整数中包含子字符串13且能被13整除的数。

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B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2668 Accepted Submission(s): 1467

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

数位DP模板题、只不过好久好久没写数位DP了、都忘了

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

int bit[15];
int dp[15][15][3];

int dfs(int pos,int s1,int s2,bool limit)  //s1代表余数，s2代表状态
{
    if(pos==-1)
    {
        return (s1==0 && s2==2);
    }
    if(!limit && dp[pos][s1][s2]!=-1) return dp[pos][s1][s2];
    int ans=0;
    int end=limit?bit[pos]:9;
    for(int i=0;i<=end;i++)
    {
        int nows2=0;
        if(s2==0)
        {
            if(i==1) nows2=1;
        }
        else if(s2==1)
        {
            if(i==1) nows2=1;
            else if(i==3) nows2=2;
        }
        else if(s2==2) nows2=2;
        ans+=dfs(pos-1,(s1*10+i)%13,nows2,limit && i==end);
    }
    if(!limit) dp[pos][s1][s2]=ans;
    return ans;
}

int cal(int n)
{
    int len=0;
    while(n)
    {
        bit[len++]=n%10;
        n/=10;
    }
    return dfs(len-1,0,0,1);
}
int main()
{
    int n;
    memset(dp,-1,sizeof(dp));
    while(scanf("%d",&n)!=EOF)
    {
        printf("%d\n",cal(n));
    }
    return 0;
}