HDU 3652 B-number

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

总算用记忆化搜索搞定了一个难一点的数位dp了！。

AC代码例如以下：

///记忆化搜素  500MS 272K

#include<iostream>
#include<cstdio>
#include<cstring>
#define mod 13
using namespace std;

int dp[20][15][4];
int num[20];

int dfs(int pos,int mo,int status,bool limit)
{
    int i;
    //cout<<pos<<"~~~~~~~"<<mo<<"~~~~~"<<status<<"~~~~~~~~"<<limit<<"~~~~~~~"<<dp[pos][status]<<endl;
    if(!pos)
        return status==2&&mo==0;
    if(!limit&&dp[pos][mo][status]!=0) return dp[pos][mo][status];
    int end = limit ?
 num[pos] : 9;
    int sum=0;
    for(i=0;i<=end;i++)
    {
        int a=mo;
        int flag = status ;
        if(flag==0&&i==1) flag=1;
        if(flag==1&&i==3) flag=2;
        if(flag==1&&i!=1&&i!=3) flag=0;
        sum+=dfs(pos-1,(a*10+i)%mod,flag,limit&&i==end);
    }
    //cout<<"!!!!!!!!!!!"<<sum<<"!!!!!!!!"<<endl;
    return limit ?
 sum : dp[pos][mo][status] = sum;
}

int _13(int n)
{
    int pos=1;
    memset(dp,0,sizeof dp);
    while (n>0)
    {
        num[pos++]=n%10;
        n/=10;
    }
    //cout<<"~~~~~~~~~"<<pos-1<<"~~~~~~~~"<<endl;
    return dfs(pos-1,0,0,true);
}

int main()
{
    int i,j;
    int n;
    while(~scanf("%d",&n))
    {
        printf("%d\n",_13(n));
    }
    return 0;
}