結果
| 問題 |
No.1917 LCMST
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-15 23:58:16 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,367 bytes |
| コンパイル時間 | 399 ms |
| コンパイル使用メモリ | 81,664 KB |
| 実行使用メモリ | 239,340 KB |
| 最終ジャッジ日時 | 2025-04-15 23:59:43 |
| 合計ジャッジ時間 | 15,264 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 9 WA * 33 |
ソースコード
import sys
import math
def main():
input = sys.stdin.read().split()
n = int(input[0])
A = list(map(int, input[1:n+1]))
if n == 1:
print(0)
return
# Find m1, count_m1, m2, count_m2
m1 = float('inf')
m2 = float('inf')
count_m1 = 0
count_m2 = 0
for a in A:
if a < m1:
m2 = m1
count_m2 = count_m1
m1 = a
count_m1 = 1
elif a == m1:
count_m1 += 1
elif a < m2:
m2 = a
count_m2 = 1
elif a == m2:
count_m2 += 1
if count_m1 >= 2:
# Case 1: multiple m1, connect all to m1
sum_lcm = 0
for a in A:
g = math.gcd(m1, a)
sum_lcm += (m1 * a) // g
sum_result = sum_lcm - m1
print(sum_result)
else:
# Case 2: connect m1 to m2, then others to m1 or m2
g = math.gcd(m1, m2)
lcm_m1m2 = (m1 * m2) // g
sum_total = lcm_m1m2 + (count_m2 - 1) * m2
for a in A:
if a == m1 or a == m2:
continue
g1 = math.gcd(m1, a)
lcm1 = (m1 * a) // g1
g2 = math.gcd(m2, a)
lcm2 = (m2 * a) // g2
sum_total += min(lcm1, lcm2)
print(sum_total)
if __name__ == '__main__':
main()
lam6er