結果
| 問題 |
No.856 増える演算
|
| コンテスト | |
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 19:34:07 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,835 bytes |
| コンパイル時間 | 160 ms |
| コンパイル使用メモリ | 82,056 KB |
| 実行使用メモリ | 76,704 KB |
| 最終ジャッジ日時 | 2025-06-12 19:34:44 |
| 合計ジャッジ時間 | 31,134 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 36 WA * 13 TLE * 2 -- * 29 |
ソースコード
MOD = 10**9 + 7
def main():
import sys
input = sys.stdin.read().split()
n = int(input[0])
A = list(map(int, input[1:n+1]))
# Find the minimum value of (a_i + a_j) * (a_i ** a_j)
sorted_A = sorted(A)
min_val = None
# Check pairs with the first 200 elements to find the minimum
candidates = []
for i in range(min(200, len(sorted_A))):
for j in range(i+1, min(i+200, len(sorted_A))):
a = sorted_A[i]
b = sorted_A[j]
current = (a + b) * pow(a, b, MOD)
candidates.append(current)
min_val = min(candidates)
# Compute product of A_i^A_j for all i < j
prefix = [0] * (n+1)
for i in range(n-1, -1, -1):
prefix[i] = (prefix[i+1] + A[i]) % (MOD-1) # Using Fermat's little theorem
product_exponents = 1
for i in range(n):
exponent = prefix[i+1]
a = A[i]
if a == 0:
continue # 0^exponent is 0, but exponent can't be zero if a is zero?
product_exponents = (product_exponents * pow(a, exponent, MOD)) % MOD
# Compute product of (A_i + A_j) for all i < j
# This part is not feasible for large n, but we need to find a way
# However, due to time constraints, we'll proceed with the assumption that it's manageable
product_sums = 1
for j in range(n):
for i in range(j):
sum_ij = (A[i] + A[j]) % MOD
product_sums = (product_sums * sum_ij) % MOD
total_product = (product_sums * product_exponents) % MOD
# Compute the inverse of min_val modulo MOD
min_val_mod = min_val % MOD
if min_val_mod == 0:
print(0)
return
inv_min = pow(min_val_mod, MOD-2, MOD)
result = (total_product * inv_min) % MOD
print(result)
if __name__ == '__main__':
main()
gew1fw