HDOJ 3652 B-number

 

B-number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1561    Accepted Submission(s): 854

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

 

加一个参数记录%13的余数。。。。newres=(res×10+i)%13


 

#include <iostream> #include <cstdio> #include <cstring> using  namespace std; int dp[ 20][ 20][ 3],n,bit[ 20],len; /*     dp『位』『余数』『状态』     0--->没有13     1--->没有13但是前面一位是1     2--->有13 */ int dfs( int pos, int res, int s, bool limit) {      if(pos==- 1)          return (s== 2)&&(res== 0);      if(!limit&&~dp[pos][res][s])          return dp[pos][res][s];      int end=limit?bit[pos]: 9;      int ans= 0;      for( int i= 0;i<=end;i++)     {          int newres=(res* 10+i)% 13;          int news=s;          if(s== 0&&i== 1)             news= 1;          else  if(s== 1&&i== 3)             news= 2;          else  if(s== 1&&i== 1)             news= 1;          else  if(s== 1&&i!= 1)             news= 0;         ans+=dfs(pos- 1,newres,news,limit&&i==end);     }      if(!limit)         dp[pos][res][s]=ans;      return ans; } int main() {     memset(dp,- 1, sizeof(dp));      while(scanf( "%d",&n)!=EOF)     {         len= 0;          while(n)         {             bit[len++]=n% 10;             n/= 10;         }         printf( "%d\n",dfs(len- 1, 0, 0, true));     }      return  0; }

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