687B: Remainders Game

Codeforces Round #360 Editorial [+ Challenges!]

B. Remainders Game

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
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.

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

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.

Hint

Assume the answer of a test is No. There must exist a pair of integers
x1 and
x2 such that both of them have the same remainders after dividing by any
ci, but they differ in remainders after dividing by
k. Find more facts about
x1 and x2!

Solution

Consider the x1 and
x2 from the hint part. We have
x1 - x2 ≡ 0 (

)
for each 1 ≤ i ≤ n.

So:



We also have

(

).
As a result:



We've found a necessary condition. And I have to tell you it's also sufficient!

Assume

, we are going to prove there exists
x1, x2 such that
x1 - x2 ≡ 0 (

)
(for each 1 ≤ i ≤ n), and


(

).

A possible solution is x1 = lcm(c1, c2, ..., cn)
and x2 = 2 × lcm(c1, c2, ..., cn), so the sufficiency
is also proved.

So you have to check if lcm(c1, c2, ..., cn) is divisible by
k, which could be done using prime factorization of
k and ci values.

For each integer x smaller than
MAXC, find it's greatest prime divisor gpdx using
sieve of Eratosthenes in


.

Then using gpd array, you can write the value of each coin as
p1q1p2q2...pmqm
where pi is a prime integer and
1 ≤ qi holds. This could be done in


by moving from
ci to


and adding
gpdci to the answer. And you can factorize
k by the same way. Now for every prime
p that

, see if there exists any coin
i that the power of
p in the factorization of ci is not smaller than the power of
p in the factorization of
k.

Complexity is

.

C++ code

//     . .. ... .... ..... be name khoda ..... .... ... .. .     \\

#include <bits/stdc++.h>
using namespace std;

inline int in() { int x; scanf("%d", &x); return x; }
const long long N = 1200021;

int cntP
, isP
;

int main()
{
for(int i = 2; i < N; i++)
if(!isP[i])
for(int j = i; j < N; j += i)
isP[j] = i;
int n = in(), k = in();
for(int i = 0; i < n; i++)
{
int x = in();
while(x > 1)
{
int p = isP[x];
int cnt = 0;
while(x % p == 0)
{
cnt++;
x /= p;
}
cntP[p] = max(cntP[p], cnt);
}
}
bool ok = 1;
while(k > 1)
{
ok &= (cntP[isP[k]] > 0);
cntP[isP[k]]--;
k /= isP[k];
}
cout << (ok ? "Yes\n" : "No\n");
}

原文链接：Codeforces Round #360 Editorial [+ Challenges!] - Codeforces