Description

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 Input

1 10 2

2  4  5

Sample Output

1 2 4 8

-1

此题非常要注意什么时候循环结束,否则可能会爆long long

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
using namespace std;
LL powll(LL x, LL n)
{
LL pw = 1;
while (n > 0)
{
if (n & 1)
pw *= x;
x *= x;
n >>= 1;
}
return pw;
}
int main()
{
LL l,r,k;
LL x;
cin>>l>>r>>k;
int flag=0;
for(int i=0;i<=66;i++)
{
if(x>r/k) break;
x=powll(k,i);
if(x>=l&&x<=r)
{
flag=1;
cout<<x<<" ";
}
}
if(flag==0)
{
puts("-1");
}
return 0;
}

  

Codeforces Round #339 (Div. 2) A的更多相关文章

  1. Codeforces Round #339 (Div.2)

    A. Link/Cut Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...

  2. Codeforces Round #339 (Div. 1) A. Peter and Snow Blower 计算几何

    A. Peter and Snow Blower 题目连接: http://www.codeforces.com/contest/613/problem/A Description Peter got ...

  3. Codeforces Round #339 (Div. 2) B. Gena's Code 水题

    B. Gena's Code 题目连接: http://www.codeforces.com/contest/614/problem/B Description It's the year 4527 ...

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

    A. Link/Cut Tree 题目连接: http://www.codeforces.com/contest/614/problem/A Description Programmer Rostis ...

  5. Codeforces Round #339 (Div. 1) C. Necklace 构造题

    C. Necklace 题目连接: http://www.codeforces.com/contest/613/problem/C Description Ivan wants to make a n ...

  6. Codeforces Round #339 (Div. 1) B. Skills 暴力 二分

    B. Skills 题目连接: http://www.codeforces.com/contest/613/problem/B Description Lesha plays the recently ...

  7. Codeforces Round #339 Div.2 C - Peter and Snow Blower

    Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. A ...

  8. Codeforces Round #339 Div.2 B - Gena's Code

    It's the year 4527 and the tanks game that we all know and love still exists. There also exists Grea ...

  9. Codeforces Round #339 Div.2 A - Link/Cut Tree

    第一次正式参加常规赛想想有些小激动的呢 然后第一题就被hack了 心痛 _(:зゝ∠)_ tle点在于越界 因此结束循环条件从乘变为除 done //等等 这题没过总评 让我静静........ // ...

随机推荐

  1. ava的打包jar、war、ear包的作用、区别、打包方式

    编为大家介绍,基于Java的打包jar.war.ear包的作用与区别详解.需要的朋友参考下以最终客户的角度来看,JAR文件就是一种封装,他们不需要知道jar文件中有多少个.class文件,每个文件中的 ...

  2. HDU 6397(2018多校第8场1001) Character Encoding 容斥

    听了杜教的直播后知道了怎么做,有两种方法,一种构造函数(现在太菜了,听不懂,以后再补),一种容斥原理. 知识补充1:若x1,x2,.....xn均大于等于0,则x1+x2+...+xn=k的方案数是C ...

  3. C++字符串流保存数据

    文件流是以外存文件为输入输出对象的数据流.字符串流是以内存中用户定义的字符数组(字符串)为输入输出对象的. 建立输出字符串流: ostrstream strout(c,sizeof(c));第一个参数 ...

  4. boost::fucntion 用法详解

    转载自:http://blog.csdn.net/benny5609/article/details/2324474 要开始使用 Boost.Function, 就要包含头文件 "boost ...

  5. Luogu 3939 数颜色

    随手点开一个题. 咦,这不是裸的动态开点线段树吗?写一个写一个…… Code: #include <cstdio> #include <cstring> using names ...

  6. 第三章:PCL基础3.1

    架构师为了确保在PCL中所有代码风格的一致性,使得其他开发者及用户容易理解源码,PCL开发者制定并遵循着一套严格的编写规范,PCL的开发者都默认此规范. 3.1PCL推荐的命名规范 1.文件命名 1) ...

  7. 【IMOOC学习笔记】多种多样的App主界面Tab实现方法(二)

    Fragment实现Tab 首先把activity_main.xml 文件中的ViewPager标签改成Fragment标签 <FrameLayout android:id="@+id ...

  8. jQuery 插件开发——GridData(表格)

    导读:我个人认为做开发最幸福的事之一就是设计一套属于自己的控件,老早之前就想去做这样的事情,一直碍于事件的冲突和个人的想法,最终没有定论,最近难得抽出一些空隙,去完成这件事情.其实自定义控件并不是难事 ...

  9. DropDownList判断值是否存在下拉列表中

    //1.值是text string aa= Request.QueryString["CallReason"].ToString();//获取传值 if (DropDownList ...

  10. sqlserver快速删除整个表数据

    --删除整个表数据 SET STATISTICS TIME ON; DECLARE @Timer DATETIME = GETDATE(); TRUNCATE TABLE LOG_DEBUG_ERRO ...