CodeForces 614A Link/Cut Tree

最新推荐文章于 2022-09-18 16:32:08 发布

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最新推荐文章于 2022-09-18 16:32:08 发布

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本文详细介绍了如何解决在给定区间内找到所有k的幂的问题，通过避免溢出和利用数学技巧来高效实现。重点在于理解幂的计算边界和优化算法。

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Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the expose procedure.

Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)

Given integers l, r and k, you need to print all powers of number k within range from l to r inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!

Input

The first line of the input contains three space-separated integers l, r and k (1 ≤ l ≤ r ≤ 10¹⁸, 2 ≤ k ≤ 10⁹).

Output

Print all powers of number k, that lie within range from l to r in the increasing order. If there are no such numbers, print "-1" (without the quotes).

Sample test(s)

Input

1 10 2

Output

1 2 4 8

Input

2 4 5

Output

-1

Note

Note to the first sample: numbers 2⁰ = 1, 2¹ = 2, 2² = 4, 2³ = 8 lie within the specified range. The number 2⁴ = 16 is greater then 10, thus it shouldn't be printed.

简单水题，不过有坑

给一个区间l,r.问在这个区间里为k的幂的数是哪些。

方法很简单，就是暴力枚举。

但是k的幂值在中间过程中可能会超long long，溢出，在判断时会出现意想不到的错误。

这题C++WA了之后果断Java大数。

import java.util.Scanner;
import static java.lang.System.*;
import java.math.BigInteger;

public class Main {

	public static void main(String[] args) {
		Scanner jin=new Scanner(in);
		long l,r,k,left,right;
		boolean isSolve=false;
		l=jin.nextLong();
		r=jin.nextLong();
		k=jin.nextLong();
		left=(long)Math.floor(Math.log(l)/Math.log(k));
		right=(long)Math.ceil(Math.log(r)/Math.log(k));
		BigInteger ans=BigInteger.valueOf((long)Math.pow(k,left));
		for(long i=left;i<=right;i++){
			if(ans.compareTo(BigInteger.valueOf(l))==-1){
				ans=ans.multiply(BigInteger.valueOf(k));
				continue;
			}
			if(ans.compareTo(BigInteger.valueOf(r))==1)
				break;
			isSolve=true;
			out.print(ans);
			out.print(' ');
			ans=ans.multiply(BigInteger.valueOf(k));
		}
		if(!isSolve)
			out.print("-1");
		out.println();
	}
	
}