Codeforces-687B Remainders Game

最新推荐文章于 2024-05-06 23:17:47 发布

_Wilbert

最新推荐文章于 2024-05-06 23:17:47 发布

阅读量794

点赞数

分类专栏： Codeforce 文章标签：数论中国剩余定理

本文链接：https://blog.youkuaiyun.com/Snow_Me/article/details/51800346

版权

Codeforce 专栏收录该内容

21 篇文章

订阅专栏

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 $\text{[math]}$ . There are n ancient numbers c1, c2, ..., cn and Pari has to tell Arya $\text{[math]}$ 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 $\text{[math]}$ for any positive integer x?

Note, that $\text{[math]}$ means the remainder of x after dividing it by y.

Input

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).

Output

Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.

   Examples 
 
     input 
   
4 5
2 3 5 12

     output 
   
Yes

     input 
   
2 7
2 3

     output 
   
No

Note

In the first sample, Arya can understand $\text{[math]}$ because 5 is one of the ancient numbers.

In the second sample, Arya can't be sure what $\text{[math]}$ 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.

题目大意：

给你n个数ci，和一个k，已知x mod ci，问你能不能确定x mod k

解法：

等于给你n个数，让你求x，然后确定x mod k是不是同一个数

中国剩余定理，唔贴个链接，这个博客说的反正我能看懂。。。

http://blog.youkuaiyun.com/acdreamers/article/details/8050018

然后其实最后我们得到的结论就是只要k 能整除 ci的最小公倍数那么就可以得到x mod k确定

以下是代码。

#include <cstdio>
using namespace std;
typedef long long LL;
LL gcd(LL a, LL b){
	while(b){
		int tmp = a % b;
		a = b;
		b = tmp;
	}
	return a;
}
int main()
{
	LL n, k, c, lcm = 1;
	scanf("%I64d%I64d", &n, &k);
	for(int i = 0; i < n; ++i){
		scanf("%I64d", &c);
		
		lcm = (c * lcm) / gcd(lcm, c);
		lcm %= k;
	}
	if(lcm == 0) printf("Yes\n");
	else printf("No\n");
	return 0;
}