HDU 3652 B-number

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

 

题意：给你一个数n，要你求出从1到n中有多少个数满足能被13整除且含有子串“13”

思路：这个题目如果多数位DP理解深刻那么就很容易了，我们发现除了位数我们需要考虑，那么不同的状态有影响的是该数能否被13整除，是否含有子串13，该数的尾数

所以AC代码：

 

#include
#include
#include
using namespace std;

int f[20][20][3][10],bits[20];

int dfs(int pos,int flag,int val,int e,bool bianjie)
{
    int ans=0;
    if(pos==-1)
    {
        if(flag!=1)
            return 0;
        else
        {
            if(val==0)
            {
                return 1;
            }
            else
                return 0;
        }
    }
    if(!bianjie&&f[pos][val][flag][e]!=-1)
        return f[pos][val][flag][e];
    int u=bianjie?bits[pos]:9;
    for(int i=0;i<=u;i++)
    {
        ans+=dfs(pos-1,(e==1&&i==3)||flag,(val*10+i)%13,i,bianjie&&i==u);
    }
    return bianjie?ans:f[pos][val][flag][e]=ans;
}

int n;

int solve()
{

    int len_m=0;
    while(n)
    {
        bits[len_m++]=n%10;
        n/=10;
    }
    return dfs(len_m-1,0,0,0,true);
}


int main()
{
    memset(f,-1,sizeof(f));
    while(~scanf("%d",&n))
    {
        cout<