hdu3652 B-number（数位dp）

题目传送门

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8970    Accepted Submission(s): 5331

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

 

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题意：求[1,n]中数位有13并且能被13整除的数

题解：数位dp，状态就是要记录上一位，还要记录这个数，可开3维或者4维

代码：

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
int n;
//int dp[10][2][13];
int dp[10][10][2][13];//i:处理的数位，j:结尾的数 k:是否已经包含13 z:该数对13取模以后的值，
int bit[10];
int dfs(int  pos,int pre,bool flag,int sta,bool limit)
{
    if(pos==-1)return sta%13==0&&flag;
    if(!limit&&dp[pos][pre][flag][sta%13]!=-1)return dp[pos][pre][flag][sta%13];
    int up=limit?bit[pos]:9;
    int ans=0;
    for(int i=0;i<=up;i++)
    {
        /*if(pre==2||pre==1&&i==3)
            ans+=dfs(pos-1,2,sta*10+i,limit&&i==up);
         else if(i==1)
           ans+=dfs(pos-1,1,sta*10+i,limit&&i==up);
         else
           ans+=dfs(pos-1,0,sta*10+i,limit&&i==up);*/
           ans+=dfs(pos-1,i,flag||(pre==1&&i==3),sta*10+i,limit&&i==up);
    }
    if(!limit)dp[pos][pre][flag][sta%13]=ans;
     return ans;
}
int calc(int x)
{
    int len=0;
    while(x)
    {
        bit[len++]=x%10;
         x/=10;
    }
    return dfs(len-1,-1,0,0,1);
}
int main()
{
    memset(dp,-1,sizeof(dp));
    while(~scanf("%d",&n))
    {
       printf("%d
",calc(n));
    }
    return 0;
}