HDU 3652 B-number（数位DP）

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1231    Accepted Submission(s): 651

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

 

 

比较简单的数位DP，随便搞

/*

 * HDU 3652 B-number

 * 含有数字13和能够被13整除的数的个数

 * dp[i][j][k][z]：i:处理的数位，j:该数对13取模以后的值，k:是否已经包含13,z结尾的数

 */

#include <iostream>

#include <string.h>

#include <algorithm>

#include <stdio.h>

using namespace std;

int dp[12][15][2][10];

int bit[12];

int dfs(int pos,int num,bool t,int e,bool flag)

{

    if(pos==-1)return t&&(num==0);

    if(!flag && dp[pos][num][t][e]!=-1)

        return dp[pos][num][t][e];

    int end=flag?bit[pos]:9;

    int ans=0;

    for(int i=0;i<=end;i++)

        ans+=dfs(pos-1,(num*10+i)%13,t||(e==1&&i==3),i,flag&&(i==end));

    if(!flag)dp[pos][num][t][e]=ans;

    return ans;

}

int calc(int n)

{

    int pos=0;

    while(n)

    {

        bit[pos++]=n%10;

        n/=10;

    }

    return dfs(pos-1,0,0,0,1);

}

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int n;

    memset(dp,-1,sizeof(dp));

    while(scanf("%d",&n)==1)

    {

        printf("%d\n",calc(n));

    }

    return 0;

}