HDU 4577 X-Boxes

X-Boxes

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 202    Accepted Submission(s): 71


Problem Description
Crazygirl is a crazy fan of XBOX games. Today, she’s here middle in a competition, in which the winner will be rewarded with an opportunity of working in the XBOX Company as a game testing player. Now, here comes the final game. As Cazygirl get a draw with the other competitor, Lich King, she must beat Lich this time.
The game is quite simple. There are n balls numbered from 1 to n and k boxes B1, B2,…, Bk satisfying following conditions:
1.  With any ball x in box Bi, there must be ball 2x in box Bi+1 if there is a box Bi+1;
2.  With any ball x in box Bi, there must be ball y in box Bi-1 satisfying 2y=x if there is a box Bi-1;
3.  You can’t put a ball in two different boxes at the same time;
4.  Your score is the number of balls in box B1;
5.  The player who get the highest score win the game of course.
So, you should tell Crazygirl the highest score she can get.

 

 

Input
The first line is the number of test cases.
Each test case has one line containing two integers n and k, meaning that there are n balls and k boxes. ( 1≤n≤1010000, 2≤k≤25 )
 

 

Output
For each test case, output one line that contains an integer equals to the highest score you can get.
 

 

Sample Input
3 10 2 7 5 8 3
 

 

Sample Output
4 0 1
 

 

Source
 
 
 
首先将所有数进行分组:
1 2 4 8 16 ......
3 6 12 24 48.....
5 10 20 40 80.....
.....
 
 
 
其实就是奇数的2^n倍。
 
 
 
可以放在1的相当于是(2*a-1)*2^(k*t-1) <= n
 
 
然后枚举t,求多少个a满足。‘’
 
 
这题很卡时间,C++写的大数TLE到现在。
 
JAVA一写就过了,以后多学着使用JAVA了,好方便
 
 
 
 1 import java.io.*;
 2 import java.math.*;
 3 import java.util.*;
 4 
 5 public class Main 
 6 {
 7     public static void main(String[] args)
 8     {
 9         Scanner cin = new Scanner(new BufferedInputStream(System.in));
10         BigInteger n;
11         int k;
12         int T;
13         T = cin.nextInt();
14         while(T>0)
15         {
16             T--;
17             n = cin.nextBigInteger();
18             k = cin.nextInt();
19             int tmp = (1<<k);
20             BigInteger ans = BigInteger.ZERO;
21             n = n.multiply(BigInteger.valueOf(2));
22             while(n.compareTo(BigInteger.ZERO) > 0)
23             {
24                 n = n.divide(BigInteger.valueOf(tmp));
25                 BigInteger tt = n.add(BigInteger.ONE).divide(BigInteger.valueOf(2));
26                 ans = ans.add(tt);
27             }
28             System.out.println(ans);
29         }
30     }
31 }

 

 
 
 
 
 
 
 
 

 

 

 

posted on 2013-08-11 15:30 kuangbin 阅读(...) 评论(...) 编辑 收藏

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