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 B ₁, B ₂,…, B _k satisfying following conditions:
1.  With any ball x in box B _i, there must be ball 2x in box B _i+1 if there is a box B _i+1;
2.  With any ball x in box B _i, there must be ball y in box B _i-1 satisfying 2y=x if there is a box B _i-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 B ₁;
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≤10 ¹⁰⁰⁰⁰, 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

2013ACM-ICPC杭州赛区全国邀请赛

 

 

 

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 }