HDU 3652 B-number (数位dp)

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本文介绍了一种用于计算特定范围内所有B-number数量的算法。B-number是指非负整数中包含子串'13'且能被13整除的数。通过递归状态压缩动态规划方法实现，有效地解决了该问题。

B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7409 Accepted Submission(s): 4340

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

Sample Output

#include <stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[20];
int dp[20][20][3];//sta 为判断余数是否为0   sta1 0 不为1  1为1 2为 13 
int dfs(int pos,int mod,int have,bool limit)//sta 为 mod sta1为数位判断 
{
	if(pos==-1)
	return mod==0&&have==2;
	
	if(dp[pos][mod][have]!=-1&&!limit)
	return dp[pos][mod][have];
	int up=limit?a[pos]:9;
	int sum=0;
	int have_x,i;
	for(i=0;i<=up;i++)
	{
		have_x=have;
		if(have_x==0&&i==1)
		{
			have_x=1;	
		}
		if(have_x==1&&i==3)
		{
			have_x=2;
		}	  
		else if(have_x==1&&i!=1)
		{
			have_x=0;
		}
		sum+=dfs(pos-1,(mod*10+i)%13,have_x,limit&&i==up);
	}
	if(!limit)
	{
		dp[pos][mod][have]=sum;
	}
	return sum;
}
int solve(int x)
{
	int cnt=0;
	while(x)
	{
		a[cnt++]=x%10;
		x/=10;
	}
	return dfs(cnt-1,0,0,1);
}
int main(int argc, char *argv[])
{	
	memset(dp,-1,sizeof(dp));
	memset(a,-1,sizeof(a));
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		printf("%d\n",solve(n));
	}
	return 0;
}