hdu 3652 B-number 数位dp

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

数位dp

看的dp进阶之路

 

dp[i][j][k]表示第i位j状态k表示mod13的余数

j==0表示前面没有13的；

j==1表示前面一位为1；

j==2表示前面含有13；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e2+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
ll bit[N];
ll f[N][5][13];
ll dp(int pos,ll pre,int flag,int m)
{
    if(pos==0)return (pre==2&&m==0);
    if(flag&&f[pos][pre][m]!=-1)return f[pos][pre][m];
    ll x=flag?9:bit[pos];
    ll ans=0;
    for(ll i=0;i<=x;i++)
    {
        if(pre==2||(pre==1&&i==3))
            ans+=dp(pos-1,2,flag||i<x,(m*10+i)%13);
        else if(i==1)
            ans+=dp(pos-1,1,flag||i<x,(m*10+i)%13);
        else
            ans+=dp(pos-1,0,flag||i<x,(m*10+i)%13);
    }
    return f[pos][pre][m]=ans;
}
ll getans(ll x)
{
    memset(f,-1,sizeof(f));
    int len=0;
    while(x)
    {
        bit[++len]=x%10;
        x/=10;
    }
    return dp(len,0,0,0);
}
int main()
{
    ll n;
    while(~scanf("%lld",&n))
    {
        printf("%lld
",getans(n));
    }
    return 0;
}