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): 6128    Accepted Submission(s): 3528

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

题意是求1~n里有多少个数，数位上有13存在且这个数取余13==0

数位dp

state：1,：前一位是1    2：前面有13出现    否则0

当时不知道怎么满足取余13等于0，其实按位数的权值一位一位的加，并膜13，只保留余数即可

这样：dp【位数】【state】【余数】

核心是想到记录余数这个状态，只有前面的余数相同，后面能取的个数才和之前求并记忆的一样

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define eps 1e-9
#define PI 3.141592653589793
#define bs 1000000007
#define bsize 256
#define MEM(a) memset(a,0,sizeof(a))
#include<iostream>
typedef long long ll;
using namespace std;
int dp[100][3][15];
int a[100];
int dfs(int pos,int state,int limit,int yu,long long t)
{
	if(pos==-1)
    {
        return state==2&&yu==0;
    }

	if(!limit&&dp[pos][state][yu]!=-1)
	return dp[pos][state][yu];
	int up=limit?a[pos]:9;
	int ans=0;
	int now;
	for(int i=0;i<=up;i++)
	{
        if(state==2)
            now=2;
	    else if(state==1&&i==3)
            now=2;
        else if(i==1)
            now=1;
        else
            now=0;
		ans+=dfs(pos-1,now,limit&&i==a[pos],(yu+i*t)%13,t/10);
	}
	if(!limit&&dp[pos][state][yu]==-1)
    {
        dp[pos][state][yu]=ans;
    }
    return ans;
 }
 int solve(int x)
 {
     int pos=0;
     long long t=1;
     while(x)
     {
         a[pos++]=x%10;
         x/=10;
         t*=10;
     }
     return dfs(pos-1,0,1,0,t/10);
 }
int main()
{
    int n;
    memset(dp,-1,sizeof(dp));
    while(~scanf("%d",&n))
    {
        printf("%d\n",solve(n));
    }
    return 0;
 }