Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) A. Bear and Poker 分解
A. Bear and Poker
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/573/problem/A
Description
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are n players (including Limak himself) and right now all of them have bids on the table. i-th of them has bid with size ai dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input
First line of input contains an integer n (2 ≤ n ≤ 105), the number of players.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109) — the bids of players.
Output
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Sample Input
4
75 150 75 50
Sample Output
Yes
HINT
题意
给你n个数,你可以把操作任意一个数,使他翻倍或者翻三倍无数次,问你是否可以让所有数都相同
题解:
假设最后变成了K,那么K=2^x+3^y+tmp
所以把所有数都按照2和3分解之后,只要剩下的数都是tmp,那就是yes
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200051
#define mod 10007
#define eps 1e-9
int Num;
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** ll a[maxn];
int check(ll x)
{
if(x==)
return ;
if(x==)
return ;
if(x==)
return ;
if(x%==)
return check(x/);
if(x%==)
return check(x/);
return ;
}
int n;
ll gcd(ll a,ll b)
{
return b==?a:gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
return a*b/gcd(a,b);
}
int main()
{
n=read();
for(int i=;i<n;i++)
a[i]=read();
sort(a,a+n);
ll tmp = a[];
for(int i=;i<n;i++)
{
tmp = gcd(tmp,a[i]);
}
for(int i=;i<n;i++)
{
if(!check(a[i]/tmp))
{
cout<<"No"<<endl;
return ;
}
}
cout<<"Yes"<<endl;
}
Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) A. Bear and Poker 分解的更多相关文章
- Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) B. Bear and Blocks 水题
B. Bear and Blocks Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/573/pr ...
- Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) C. Bear and Drawing
题目链接:http://codeforces.com/contest/573/problem/C题目大意:在两行无限长的点列上面画n个点以及n-1条边使得构成一棵树,并且要求边都在同一平面上且除了节点 ...
- 校内选拔I题题解 构造题 Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2) ——D
http://codeforces.com/contest/574/problem/D Bear and Blocks time limit per test 1 second memory limi ...
- Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2) A. Bear and Elections 优先队列
A. Bear and Elections ...
- Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2) B. Bear and Three Musketeers 枚举
B. Bear and Three Musketeers ...
- Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2)C. Bear and Poker
C. Bear and Poker ...
- Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2)
以后每做完一场CF,解题报告都写在一起吧 暴力||二分 A - Bear and Elections 题意:有n个候选人,第一个候选人可以贿赂其他人拿到他们的票,问最少要贿赂多少张票第一个人才能赢 ...
- Codeforces Round #318 (Div. 2) C Bear and Poker (数学)
简单题,求一下所有数的2和3的幂是任意调整的,把2和3的因子除掉以后必须相等. 求lcm,爆了long long.我得好好反省一下,对连乘不敏感 #include<bits/stdc++.h&g ...
- Codeforces Round #539Ȟȟȡ (Div. 1) 简要题解
Codeforces Round #539 (Div. 1) A. Sasha and a Bit of Relax description 给一个序列\(a_i\),求有多少长度为偶数的区间\([l ...
随机推荐
- Android 如何直播RTMP流
在android上,视频/音频流直播是极少有人关注的一部分.每当我们讨论流媒体,RTMP(Real Time Messaging Protocol)是不可或缺的.RTMP是一个基本的视频/音频直播流协 ...
- 【转】移动web资源整理
目录(更新于20150311) meta基础知识 H5页面窗口自动调整到设备宽度,并禁止用户缩放页面 忽略将页面中的数字识别为电话号码 忽略Android平台中对邮箱地址的识别 当网站添加到主屏幕快速 ...
- Jquery图片放大镜
一般在“在线商城.电子商务.企业产品介绍”等地方经常会看到一些图片放大镜的功能,而做这个功能一般是会用一个js包——enlarge.js(这是jquery图片放大镜的插件).Enlarge 是一个基于 ...
- 【c++内存分布系列】单独一个类
首先要明确类型本身是没有具体地址的,它是为了给编译器生成相应对象提供依据.只有编译器生成的对象才有明确的地址. 一.空类 形如下面的类A,类里没有任何成员变量,类的sizeof值为1. #includ ...
- UITextView 相关知识点
1.得到UITextView的高度 - (CGRect)contentSizeRectForTextView:(UITextView *)textView { [textView.layoutMana ...
- FS,FT,DFS,DTFT,DFT,FFT的联系和区别
DCT变换的原理及算法 文库介绍 对于初学数字信号处理(DSP)的人来说,这几种变换是最为头疼的,它们是数字信号处理的理论基础,贯穿整个信号的处理. 学习过<高等数学>和<信号与系统 ...
- 瞬间从IT屌丝变大神——JavaScript规范
JavaScript规范主要包含以下内容: 底层JavaScript库采用YUI 2.8.0. 统一头部中只载入YUI load组件,其它组件都通过loader对象加载. JavaScript尽量避免 ...
- openstack分布式安装
一. keystone安装笔记 初次接触openstack,在尝试过单机部署以后不是很满意,就开始着手分布式部署,主要是按照openstack官网上的安装教程来的,本人安装的是目前最新的 I 版. 以 ...
- URL的格式scheme
url scheme: scheme + (://) + userinfo + (@) + hostname + (:) + port + path + (?) + query + (#) + fra ...
- Android实例-获取安卓手机WIFI信息(XE8+小米2)
结果: 1.必须打开Access wifi state权限,不打开权限会出图二的错误. 相关资料: http://blog.csdn.net/lyf_lyf/article/category/1735 ...