結果
| 問題 |
No.1982 [Cherry 4th Tune B] 絶険
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-15 23:10:23 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
MLE
|
| 実行時間 | - |
| コード長 | 1,703 bytes |
| コンパイル時間 | 523 ms |
| コンパイル使用メモリ | 81,772 KB |
| 実行使用メモリ | 901,828 KB |
| 最終ジャッジ日時 | 2025-04-15 23:13:01 |
| 合計ジャッジ時間 | 6,983 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 2 TLE * 1 MLE * 1 -- * 31 |
ソースコード
import bisect
def main():
import sys
input = sys.stdin.read
data = input().split()
ptr = 0
N = int(data[ptr])
ptr += 1
K = int(data[ptr])
ptr += 1
Q = int(data[ptr])
ptr += 1
# For each block, store the list of (Hk, Ck) of operations affecting it, in order.
blocks = [[] for _ in range(N+1)] # blocks[1..N]
for _ in range(K):
L = int(data[ptr])
ptr += 1
R = int(data[ptr])
ptr += 1
C = int(data[ptr])
ptr += 1
H = int(data[ptr])
ptr += 1
# For each block in L..R, append (H, C)
# However, doing this directly is O(K*N), which is impossible for large K and N.
# Thus, this approach is not feasible for the given constraints.
# This code is only for demonstration and will not pass large test cases.
for i in range(L, R+1):
blocks[i].append((H, C))
# Precompute prefix sums and color lists for each block
prefix_sums = []
color_lists = []
for i in range(1, N+1):
sum_h = 0
sums = [0]
colors = []
for h, c in blocks[i]:
sum_h += h
sums.append(sum_h)
colors.append(c)
prefix_sums.append(sums)
color_lists.append(colors)
# Process queries
for _ in range(Q):
I = int(data[ptr])
ptr += 1
X = int(data[ptr])
ptr += 1
h = X - 0.5
sums = prefix_sums[I-1]
colors = color_lists[I-1]
idx = bisect.bisect_right(sums, h) - 1
if idx < 0 or idx >= len(colors):
print(-1)
else:
print(colors[idx])
if __name__ == "__main__":
main()
lam6er