HDOJ 3652 B-number

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1561    Accepted Submission(s): 854

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

 

Recommend

lcy

 

加一个参数记录%13的余数。。。。newres=(res×10+i)%13

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int dp[20][20][3],n,bit[20],len;
/*
    dp『位』『余数』『状态』
    0--->没有13
    1--->没有13但是前面一位是1
    2--->有13
*/
int dfs(int pos,int res,int s,bool limit)
{
    if(pos==-1)
        return (s==2)&&(res==0);
    if(!limit&&~dp[pos][res][s])
        return dp[pos][res][s];
    int end=limit?bit[pos]:9;
    int ans=0;
    for(int i=0;i<=end;i++)
    {
        int newres=(res*10+i)%13;
        int news=s;
        if(s==0&&i==1)
            news=1;
        else if(s==1&&i==3)
            news=2;
        else if(s==1&&i==1)
            news=1;
        else if(s==1&&i!=1)
            news=0;
        ans+=dfs(pos-1,newres,news,limit&&i==end);
    }
    if(!limit)
        dp[pos][res][s]=ans;
    return ans;
}

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

* This source code was highlighted by YcdoiT. ( style: Pastie )