B-number - HDU 3652 数位dp

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本文介绍了一个特定数学问题——B-number的定义及求解方法。B-number是指非负整数中包含子串'13'且能被13整除的数。文章详细展示了通过深度优先搜索（DFS）算法进行高效计算的过程。

B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3461 Accepted Submission(s): 1942

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

Sample Output

题意：求出现13并且能被13整除的数字有多少个。

思路：dfs注意记录状态即可。主要还是看代码吧。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int T,t,n,m;
int num[20];
int dp[20][13][2][2][2][2];
int dfs(int pos,int left,bool th,bool one,bool th2,bool f)
{
    if(pos==0)
      return left==0 && th && th2;
    if(dp[pos][left][th][one][th2][f]>=0)
      return dp[pos][left][th][one][th2][f];
    int i,j,k,r,a,b,c,d,f2;
    dp[pos][left][th][one][th2][f]=0;
    if(f)
      r=num[pos];
    else
      r=9;
    for(i=0;i<=r;i++)
    {
        if(f && i==r)
          f2=1;
        else
          f2=0;
        a=(left*10+i)%13;
        if((left!=0 && a==0)||th)
          b=1;
        else
          b=0;
        if(i==1)
          c=1;
        else
          c=0;
        if((one && i==3)|| th2)
          d=1;
        else
          d=0;

        dp[pos][left][th][one][th2][f]+=dfs(pos-1,a,b,c,d,f2);
    }
    return dp[pos][left][th][one][th2][f];
}
int solve()
{
    int i,j,k,len=0;
    m=n;
    while(m)
    {
         num[++len]=m%10;
         m/=10;
    }
    memset(dp,-1,sizeof(dp));
    return dfs(len,0,0,0,0,1);
}
int main()
{
    int i,j,k;
    while(~scanf("%d",&n))
      printf("%d\n",solve());
}