codeforces 360 D - Remainders Game

原题：

Description

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, ..., c_n 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.

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, ..., c_n (1 ≤ c_i ≤ 1 000 000).

Output

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

Sample Input

Input

4 5
2 3 5 12

Output

Yes

Input

2 7
2 3

Output

No


提示： 
如下是引用http://blog.csdn.net/aozil_yang/article/details/51812908

反证法，假设解不惟一，有x1，x2，那么x1，x2知足：

x1 % ci == x2 % ci 而且x1 % k != x2 % k,因此 (x1 - x2) % ci == 0 ，(x1-x2) % k != 0

而且lcm(c1,c2,,,,cn) % ci == 0

,因此lcm % （x1-x2） == 0.

且   (x1-x2) % k != 0;

因此lcm % k != 0  

因此命题： 若是解不惟一，那么lcm % k != 0

逆否命题为：  若lcm % k == 0 那么 解惟一！

因此只须要判断lcm是否k 的整数倍，

也就是说这个问题能够转换为：

是否存在x使得  x是c1，c2，c3，，，cn，k的整数倍！

是的话就是yes，不然就是no！

 

 

代码：

#include<cstdio> #include<cstring> #include<algorithm>
using namespace std; typedef long long ll; 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() { // printf("%d\n",gcd(6,13));
    int n; ll k; scanf("%d%I64d",&n,&k); ll ans = 1; bool ok = false; for (int i = 0; i < n; ++i) { ll x; scanf("%I64d",&x); if (ans) ans = lcm(ans,x) % k; if (ans == 0)ok=true; } printf("%s\n",ok? "Yes" : "No" ); return 0; }