Problem:


Score : 500 points

Problem Statement

We have N integers. The i-th number is A_i.

\{A_i\} is said to be pairwise coprime when GCD(A_i,A_j)=1 holds for every pair (i, j) such that 1\leq i < j \leq N.

\{A_i\} is said to be setwise coprime when \{A_i\} is not pairwise coprime but GCD(A_1,\ldots,A_N)=1.

Determine if \{A_i\} is pairwise coprime, setwise coprime, or neither.

Here, GCD(\ldots) denotes greatest common divisor.

Constraints

2 \leq N \leq 10^6

1 \leq A_i\leq 10^6

Input

Input is given from Standard Input in the following format:

N
A_1 \ldots A_N

Output

If \{A_i\} is pairwise coprime, print pairwise coprime; if \{A_i\} is setwise coprime, print setwise coprime; if neither, print not coprime.

Sample Input 1

3
3 4 5

Sample Output 1

pairwise coprime

GCD(3,4)=GCD(3,5)=GCD(4,5)=1, so they are pairwise coprime.

Sample Input 2

3
6 10 15

Sample Output 2

setwise coprime

Since GCD(6,10)=2, they are not pairwise coprime. However, since GCD(6,10,15)=1, they are setwise coprime.

Sample Input 3

3
6 10 16

Sample Output 3

not coprime

GCD(6,10,16)=2, so they are neither pairwise coprime nor setwise coprime.



Sample input:
3
3 4 5


Sample output:
pairwise coprime

Submission status: Accepted

Solution code:
import base64
import subprocess


exe_bin = 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exe_bin = base64.b85decode(exe_bin)

with open("./kyomu", 'wb') as f:
    f.write(exe_bin)

subprocess.run(["chmod +x ./kyomu"], shell=True)
subprocess.run(["./kyomu"], shell=True)

Submission status: Accepted

Solution code:
import sys
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))  #空白あり


N = I()
A = LI()


from math import gcd

g = 0  # Aiたちの最大公約数
for i in range(N):
    g = gcd(g,A[i])

if g != 1:
    print('not coprime')
    exit()


B = [0]*(10**6+1)
for i in range(N):
    B[A[i]] += 1

for d in range(2,5*10**5+1):
    a = sum(B[i] for i in range(d,10**6+1,d))
    if a >= 2:
        print('setwise coprime')
        break
else:
    print('pairwise coprime')

# 計算量 O(A*logA)
# A = max(Ai)

Submission status: Accepted

Solution code:
import sys, math
input= sys.stdin.readline

N = int(input())
A_lst = list(map(int, input().split()))
lst=[0]*(10**6+1)

# 割れる最大の素数を出す
for i in range(2, 10**6+1):
  if not lst[i]:
    for j in range(i, 10**6+1, i):
      lst[j] = i
      
# not coprimeか判定する
ans = 'not coprime'
GCD = A_lst[0]
for i in range(N):
  GCD = math.gcd(GCD, A_lst[i])
  
if GCD != 1:
  print(ans)
else:
  ans = 'setwise coprime'
  post = 1
  visited = [0]*(10**6+1)
  for i in range(N):
    a = A_lst[i]
    while a > 1:
      dv = lst[a]
      if visited[dv]:
        post = 0
        break
      while a % dv == 0:
        a //= dv
      visited[dv] = 1
  
  if post:
    ans = 'pairwise coprime'
  print(ans)