Codeforces Round #360 (Div. 2) D. Remainders Game
1 second
256 megabytes
standard input
standard output
Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer x and k, and tells Arya k but not x. Arya have to find the value
. There are n ancient numbers c1, c2, ..., cn and Pari has to tell Arya
if Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or not. Formally, is it true that Arya can understand the value
for any positive integer x?
Note, that
means the remainder of x after dividing it by y.
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000 000) — the number of ancient integers and value k that is chosen by Pari.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 1 000 000).
Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.
4 5
2 3 5 12
Yes
2 7
2 3
No
In the first sample, Arya can understand
because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what
is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.
题意:给你n个数,一个k;可以告诉你xmod ci的值;求x%k是否唯一;
思路:根据中国剩余定理,如果中国剩余定理有解x,另外一个解为x+lcm(c0,c1...cn);
所以lcm%k==0;
#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+,M=1e6+,inf=1e9+;
ll gcd(ll a,ll b)
{
return b==?a:gcd(b,a%b);
}
int main()
{
ll x,y,z,i,t;
ll lcm=;
scanf("%lld%lld",&x,&y);
for(i=;i<x;i++)
{
scanf("%lld",&z);
lcm=z*lcm/gcd(z,lcm);
lcm%=y;
}
if(lcm==)
printf("Yes\n");
else
printf("No\n");
return ;
}
Codeforces Round #360 (Div. 2) D. Remainders Game的更多相关文章
- Codeforces Round #360 (Div. 2) D. Remainders Game 数学
D. Remainders Game 题目连接: http://www.codeforces.com/contest/688/problem/D Description Today Pari and ...
- Codeforces Round #360 (Div. 2) D. Remainders Game 中国剩余定理
题目链接: 题目 D. Remainders Game time limit per test 1 second memory limit per test 256 megabytes 问题描述 To ...
- Codeforces Round #360 (Div. 1) D. Dividing Kingdom II 暴力并查集
D. Dividing Kingdom II 题目连接: http://www.codeforces.com/contest/687/problem/D Description Long time a ...
- Codeforces Round #360 (Div. 2) C. NP-Hard Problem 水题
C. NP-Hard Problem 题目连接: http://www.codeforces.com/contest/688/problem/C Description Recently, Pari ...
- Codeforces Round #360 (Div. 2) B. Lovely Palindromes 水题
B. Lovely Palindromes 题目连接: http://www.codeforces.com/contest/688/problem/B Description Pari has a f ...
- Codeforces Round #360 (Div. 2) A. Opponents 水题
A. Opponents 题目连接: http://www.codeforces.com/contest/688/problem/A Description Arya has n opponents ...
- Codeforces Round #360 (Div. 1)A (二分图&dfs染色)
题目链接:http://codeforces.com/problemset/problem/687/A 题意:给出一个n个点m条边的图,分别将每条边连接的两个点放到两个集合中,输出两个集合中的点,若不 ...
- Codeforces Round #360 (Div. 1) D. Dividing Kingdom II 并查集求奇偶元环
D. Dividing Kingdom II Long time ago, there was a great kingdom and it was being ruled by The Grea ...
- Codeforces Round #360 (Div. 2) E. The Values You Can Make DP
E. The Values You Can Make Pari wants to buy an expensive chocolate from Arya. She has n coins, ...
随机推荐
- LInux进程虚拟地址空间的管理
2017-04-07 脱离物理内存的管理,今天咱们来聊聊进程虚拟内存的管理.因为进程直接分配和使用的都是虚拟内存,而物理内存则是有系统“按需”分配给进程,在进程看来,只知道虚拟内存的存在! 前言: 关 ...
- django 模板语言之 simple_tag 自定义模板
自定义函数 simple_tag a. app项目下创建templatetags目录 b. 创建任意xxoo.py文件 用做自定义py函数 c. 创建template对象 register 在函数或者 ...
- 使用selenium实现简单网络爬虫抓取MM图片
撸主听说有个网站叫他趣,里面有个社区,其中有一项叫他趣girl,撸主点进去看了下,还真不错啊,图文并茂,宅男们自己去看看就知道啦~ 接下来当然就是爬取这些妹子的图片啦,不仅仅是图片,撸主发现里面的对话 ...
- 『HTML5挑战经典』是英雄就下100层-开源讲座(一)从天而降的英雄
是英雄就下100层是一款经典的手机小游戏,以前是在诺基亚手机上十分有名.今天我们就用HTML5和lufylegend一步步地实现它. 一,准备工作 首先,你需要下载lufylegend,下载地址如下: ...
- jenkin构建项目执行脚本后,脚本中启动的进程也随之关闭的解决办法
问题描述: 之前用jenkins构建项目(maven项目)后都是通过ssh先将war文件推送到远程服务器,然后执行远程的脚本(更新项目,重启tomcat),一直没有出现问题,今天使用jenkins构建 ...
- RocketMQ 单机安装
本章快速入门指南是在本地机器上设置 RocketMQ 消息传递系统以发送和接收消息的详细说明. 在这先对RocketMQ 做一个简单介绍. RocketMQ是一个纯java.分布式.队列模型的开源消息 ...
- SMO算法精解
本文参考自:https://www.zhihu.com/question/40546280/answer/88539689 解决svm首先将原始问题转化到对偶问题,而对偶问题则是一个凸二次规划问题,理 ...
- mysql数据库从删库到跑路之mysql表操作
表介绍 表相当于文件,表中的一条记录就相当于文件的一行内容,不同的是,表中的一条记录有对应的标题,称为表的字段 id,name,qq,age称为字段,其余的,一行内容称为一条记录 内容: 1 创建表 ...
- sqlserver 2005/2008 导入超大sql文件
SQLCMD -E -dmaster -ic:\Scripts\create_db.sql 安装了Microsoft® SQL Server® 2008 R2 Native Client可用
- Git-基本操作(同SVN)
本人拜读了廖雪峰老师关于Git的讲述后整理所得 1.创建版本库: 版本库又名仓库,英文名repository,你可以简单理解成一个目录,这个目录里面的所有文件都可以被Git管理起来,每个文件的修改.删 ...