Codeforces 194B.Square
1 題目描述(題目連結)
2 題解
輸入是
n
n
n,那麼結果是:
l
c
m
(
4
∗
n
,
n
+
1
)
n
+
1
+
1
\cfrac{lcm(4*n, n+1)}{n+1} + 1
n+1lcm(4∗n,n+1)+1
即:
4
∗
n
g
c
d
(
4
∗
n
,
n
+
1
)
+
1
\cfrac{4*n}{gcd(4*n, n+1)} + 1
gcd(4∗n,n+1)4∗n+1
from math import gcd
n = int(input())
arr = input().split()
res = [4*int(i)//gcd(4*int(i), int(i) + 1) + 1 for i in arr]
for i in res:
print(i)
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