HDU3652 数位DP 记忆化搜索

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

131002001000

Sample Output

1122

 

题意：找出1~n范围内含有13并且能被13整除的数字的个数

    /***     hdu 3652  数位dp（含13且被能被13整除的数）     题目大意：求出给定区间内的数字含有“13”并且能被13整除的个数     解题思路：记忆化搜索。                dp[i][j][k][z]：i:处理的数位，j:该数对13取模以后的值，k:是否已经包含13,z结尾的数     */      #include <stdio.h>      #include <string.h>      #include <iostream>      #include <algorithm>      using namespace std;            int dp[10][15][2][10],bit[10];            int dfs(int pos,int mod,int t,int now,int flag)      {          if(pos==-1)return mod==0&&t;          if(!flag&&dp[pos][mod][t][now]!=-1)return dp[pos][mod][t][now];          int end=flag?bit[pos]:9;          int ans=0;          for(int i=0;i<=end;i++)          {              ans+=dfs(pos-1,(mod*10+i)%13,t||(now==1&&i==3),i,flag&&(i==end));          }          if(!flag)dp[pos][mod][t][now]=ans;          return ans;      }            int solve(int n)      {          int len=0;          while(n)          {              bit[len++]=n%10;              n/=10;          }          return dfs(len-1,0,0,0,1);      }            int main()      {          int n;          memset(dp,-1,sizeof(dp));          while(~scanf("%d",&n))          {              printf("%d\n",solve(n));          }          return 0;      }