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): 8160    Accepted Submission(s): 4822

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

 

#include<bits/stdc++.h>
using namespace std;
int dp[12][15][3],a[12],k[11];
int dfs(int pos,int mod,int f,int p)//位数，对13取模的值，是否有13，是否有上限
{
    if(pos==-1)
        return mod==0&&f==2;//能被13整除且含有13
    if(!p&&dp[pos][mod][f]!=-1)
        return dp[pos][mod][f];
    int num;
    if(p)num=a[pos];
    else num=9;
    int ans=0;
    for(int i=0;i<=num;i++)
    {
        int F;
        if(f==2)F=2;//已经有13
        else if(f==1&&i==3)F=2;//上一位是1且这一位是3
        else if(i==1)F=1;//这一位是1
        else F=0;//
        ans+=dfs(pos-1,(mod+i*k[pos])%13,F,p&&i==num);
    }
    if(!p)dp[pos][mod][f]=ans;
    return ans;
}
int main()
{
    int n;
    memset(dp,-1,sizeof(dp));
    k[0]=1;
    for(int i=1;i<=10;i++)
        k[i]=k[i-1]*10;
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        int pos=0;
        while(n)
        {
            a[pos++]=n%10;
            n/=10;
        }
        printf("%d\n",dfs(pos-1,0,0,1));
    }
    return 0;
}