[HDU 3652] B-number

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;

}