HDU 3652 数位DP 解题报告

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

【解题报告】
这是机房模拟ACM的A题，然而只有我一个人做了。。。
关于不能出现13的部分就非常显然了，重点就在整除上面，因为如果直接判断的话肯定是没法转移的，所以大概思路就是记录这个数除13的余数，（DP多加一维，dfs的时候加几句就可以了）。

代码如下：

#include<cstdio>  #include<cstring>  #include<algorithm>  using namespace std;  int n;  int dp[10][15][2][10],a[10];  int dfs(int pos,int mod,int t,int pre,int limit)  {      if(pos==-1)return (mod==0&&t);      if(!limit&&dp[pos][mod][t][pre]!=-1)return dp[pos][mod][t][pre];      int up=limit?a[pos]:9;      int ans=0;      for(int i=0;i<=up;++i)      {          ans+=dfs(pos-1,(mod*10+i)%13,t||(pre==1&&i==3),i,limit&&(i==up));      }      if(!limit) dp[pos][mod][t][pre]=ans;      return ans;  }  int solve(int x)  {      int len=0;      while(x)          a[len++]=x%10,x/=10;      return dfs(len-1,0,0,0,1);  }  int main()  {       memset(dp,-1,sizeof(dp));      while(scanf("%d",&n)!=EOF)      {          printf("%d\n",solve(n));      }      return 0;  }