Codeforces Round #339 (Div. 2) A. Link/Cut Tree

本文链接：https://blog.youkuaiyun.com/helloiamclh/article/details/50525602

此博客解决了一个特定问题，即在给定范围内找到所有k的幂次方数值。通过逐步计算并检查每个幂次是否落在指定区间内，从而筛选出符合条件的数。该过程涉及基本的数学运算和条件判断，旨在帮助理解和解决类似的数据结构和算法问题。

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.

题意：给你一个范围[l,r]，求k的i次方在那个范围的数

思路：看到过的人那么少，这还是A题啊，肯定有坑，写完交了一发，果然，爆longlong，所以为了防止爆，在它即将爆的时候停止就行了。。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 101000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define MINF 1<<30
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int main()
{
	ll l,r,k;
	while(scanf("%I64d%I64d%I64d",&l,&r,&k)!=EOF)
	{
		ll num=1;
		int bz=0;
		while(num<=r)
		{
			if(num>=l)
			{
				if(bz==0)
				{
					bz=1;
					printf("%I64d",num);
				}
				else
				printf(" %I64d",num);
			}
			if(num>r/k)
			break;
			num*=k;
		}
		if(bz==0)
		printf("-1\n");
		else
		printf("\n");
	}
	return 0;
}