結果
| 問題 |
No.2359 A in S ?
|
| コンテスト | |
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 20:20:10 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 2,055 bytes |
| コンパイル時間 | 186 ms |
| コンパイル使用メモリ | 82,732 KB |
| 実行使用メモリ | 138,684 KB |
| 最終ジャッジ日時 | 2025-06-12 20:20:30 |
| 合計ジャッジ時間 | 4,356 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 * 2 |
| other | AC * 2 TLE * 1 -- * 15 |
ソースコード
import bisect
def main():
import sys
input = sys.stdin.read().split()
ptr = 0
N = int(input[ptr])
ptr += 1
M = int(input[ptr])
ptr += 1
small_X = 300 # threshold for small X
small_groups = {} # key: (X, Y), value: list of (L, R) sorted by L
large_intervals = []
for _ in range(N):
L = int(input[ptr])
ptr += 1
R = int(input[ptr])
ptr += 1
X = int(input[ptr])
ptr += 1
Y = int(input[ptr])
ptr += 1
if X <= small_X:
key = (X, Y)
if key not in small_groups:
small_groups[key] = []
small_groups[key].append((L, R))
else:
large_intervals.append((X, Y, L, R))
# Preprocess small groups: sort each group's intervals by L, and prepare for binary search
for key in small_groups:
intervals = small_groups[key]
# Sort intervals by L
intervals.sort()
# Replace the group with the sorted list
small_groups[key] = intervals
# Read the queries
queries = list(map(int, input[ptr:ptr+M]))
ptr += M
for A in queries:
ans = 0
# Process small X groups
for X in range(1, small_X + 1):
Y = A % X
key = (X, Y)
if key not in small_groups:
continue
intervals = small_groups[key]
# Binary search to find the number of intervals with L <= A
# Since intervals are sorted by L, we can use bisect_right
L_list = [interval[0] for interval in intervals]
k = bisect.bisect_right(L_list, A)
# Check the first k intervals
count = 0
for i in range(k):
if intervals[i][1] >= A:
count += 1
ans += count
# Process large X intervals
for (X, Y, L, R) in large_intervals:
if A % X == Y and L <= A <= R:
ans += 1
print(ans)
if __name__ == '__main__':
main()
gew1fw