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

A. Link/Cut Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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;
}