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;
}