hdu 3652 B-number(数位DP，4级)

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1497    Accepted Submission(s): 820

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
   
   
   
   

 

Author

wqb0039

 

Source

2010 Asia Regional Chengdu Site —— Online Contest

 

Recommend

lcy

#include<cstdio>
#include<cstring>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
int dp[15][10][15][2];//位数，最后一个数，模剩数，是否有13
int bit[15],pos;
int DP(int pp,int ee,int mod,bool has,bool big)
{
  if(pp==0)return (mod==0&&has);
  if(big&&dp[pp][ee][mod][has]!=-1)
    return dp[pp][ee][mod][has];
  int ret=0;
  int kn=big?9:bit[pp];
  FOR(i,0,kn)
  {
    ret+=DP(pp-1,i,(mod*10+i)%13,has||(ee==1&&i==3),big||(bit[pp]!=i));
  }
  if(big)dp[pp][ee][mod][has]=ret;
  return ret;
}
int get(int x)
{ pos=0;
  while(x)
  {
    bit[++pos]=x%10;x/=10;
  }
  clr(dp,-1);
  return DP(pos,0,0,0,0);
}
int main()
{
  int n;
  while(cin>>n)
  {
    cout<<get(n)<<endl;
  }
}