結果

問題 No.1036 Make One With GCD 2
ユーザー terasa
提出日時 2022-06-12 14:07:30
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 753 ms / 2,000 ms
コード長 2,095 bytes
コンパイル時間 204 ms
コンパイル使用メモリ 82,588 KB
実行使用メモリ 193,700 KB
最終ジャッジ日時 2024-09-23 02:10:53
合計ジャッジ時間 21,677 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 41
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import sys
import pypyjit
import itertools
import heapq
import math
from collections import deque, defaultdict
import bisect
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 6)
pypyjit.set_param('max_unroll_recursion=-1')
def index_lt(a, x):
'return largest index s.t. A[i] < x or -1 if it does not exist'
return bisect.bisect_left(a, x) - 1
def index_le(a, x):
'return largest index s.t. A[i] <= x or -1 if it does not exist'
return bisect.bisect_right(a, x) - 1
def index_gt(a, x):
'return smallest index s.t. A[i] > x or len(a) if it does not exist'
return bisect.bisect_right(a, x)
def index_ge(a, x):
'return smallest index s.t. A[i] >= x or len(a) if it does not exist'
return bisect.bisect_left(a, x)
class SegTree:
def __init__(self, N, func, e):
self.N = N
self.func = func
self.X = [e] * (N << 1)
self.e = e
def build(self, seq):
for i in range(self.N):
self.X[self.N + i] = seq[i]
for i in range(self.N)[::-1]:
self.X[i] = self.func(self.X[i << 1], self.X[i << 1 | 1])
def add(self, i, x):
i += self.N
self.X[i] += x
while i > 1:
i >>= 1
self.X[i] = self.func(self.X[i << 1], self.X[i << 1 | 1])
def update(self, i, x):
i += self.N
self.X[i] = x
while i > 1:
i >>= 1
self.X[i] = self.func(self.X[i << 1], self.X[i << 1 | 1])
def query(self, L, R):
L += self.N
R += self.N
vL = self.e
vR = self.e
while L < R:
if L & 1:
vL = self.func(vL, self.X[L])
L += 1
if R & 1:
R -= 1
vR = self.func(self.X[R], vR)
L >>= 1
R >>= 1
return self.func(vL, vR)
N = int(input())
A = list(map(int, input().split()))
st = SegTree(N, math.gcd, 0)
st.build(A)
ans = 0
l = 0
r = 1
ans = (1 + N) * N // 2
while r <= N:
while st.query(l, r) == 1:
l += 1
ans -= r - l
r += 1
print(ans)
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