CodeForces 687B Remainders Game

数论。

如果$x$不唯一，假设存在两个解，较大的为${x_1}$，较小的为${x_2}$。

那么，

$left{ {egin{array}{*{20}{c}}
{{x_1}\% {c_i} = {x_2}\% {c_i}}\
{{x_1}\% k e {x_2}\% k}
end{array}} ight. Rightarrow left{ {egin{array}{*{20}{c}}
{({x_1} - {x_2})\% {c_i} = 0}\
{({x_1} - {x_2})\% k e 0}
end{array}} ight.$。

$∵lcm({c_1},{c_2},{c_3},......,{c_n})\% {c_i} = 0$

$∴lcm({c_1},{c_2},{c_3},......,{c_n})\% ({x_1} - {x_2}) = 0$

$∵({x_1} - {x_2})\% k e 0$

$∴lcm({c_1},{c_2},{c_3},......,{c_n})\% k e 0$

也就是说：如果解不唯一，那么$lcm({c_1},{c_2},{c_3},......,{c_n})\% k e 0$。

换句话说就是：如果解唯一，那么$lcm({c_1},{c_2},{c_3},......,{c_n})\% k = 0$。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-6;
void File()
{
    freopen("D:\in.txt","r",stdin);
    freopen("D:\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar(); x = 0;while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar();  }
}

LL gcd(LL a,LL b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}

LL lcm(LL a,LL b)
{
    return a*b/gcd(a,b);
}

int n;
LL ans=1,k;

int main()
{
    scanf("%d%lld",&n,&k);
    for(int i=1;i<=n;i++)
    {
        LL x; scanf("%lld",&x);
        ans=lcm(ans,x)%k;
    }
    if(ans) printf("No
");
    else printf("Yes
");

    return 0;
}