hdu3652B-number（数位dp）

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6444    Accepted Submission(s): 3739

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

 

 

 

题意：找出1~n范围内含有13并且能被13整除的数字的个数

 

思路:http://blog.csdn.net/libin56842/article/details/10026063

 

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

using namespace std;
int dit[15],f[15][15][3];

int dfs(int pos,int mod,int have,int lim)
{
    int num,ans,mod_x,have_x;
    if(pos<=0) return mod==0 && have==2;
    if(!lim && f[pos][mod][have]!=-1) return f[pos][mod][have];
    num=lim?dit[pos]:9;ans=0;
    for(int i=0;i<=num;i++)
    {
        mod_x=(mod*10+i)%13; have_x=have;
        if(have==0 && i==1) have_x=1;
        if(have==1 && i!=1) have_x=0;
        if(have==1 && i==3) have_x=2;
        ans+=dfs(pos-1,mod_x,have_x,lim&&i==num);
    }
    if(!lim) f[pos][mod][have]=ans;
    return ans;
}

int main()
{
    int n,len;
    while(~scanf("%d",&n))
    {
        memset(dit,0,sizeof dit);
        memset(f,-1,sizeof f);len=0;
        while(n)
        {
            dit[++len]=n%10;
            n/=10;
        }dit[len+1]=0;
        printf("%d\n",dfs(len,0,0,1));
    }
    return 0;
}