ABC279 D - Freefall

提出

TLE

code: python

import math

a, b = list(map(int, input().split()))

# 減少 -> 増加 or -> 増加

# B * n + A / √n+1 0 <= n <=

n = 1

res = a

while(True):

tmp = b * n + a / math.sqrt(n+1)

if (res < tmp):

print(res)

exit()

else:

res = tmp

n += 1

解答

code: python

import math

a, b = map(int, input().split())

n = math.floor((a / (2.0 * b)) ** (2.0 / 3.0)) - 1.0

if n < 0:

n = 0

# nは近似だから+1も考える

result = b * n + a / math.sqrt(n + 1.0)

result = min(result, b * (n + 1.0) + a / math.sqrt(n + 2.0))

print(result)

code: python

import math

a, b = map(int, input().split())

def f(n, a, b):

return a / math.sqrt(n + 1) + b * n

l, r = 0, pow(10, 18)

# ------------

# l m1 m2 r

while r - l > 2: # 3分割の幅

# ex) l = 0 r = 30：

# m1 = (0*2 + 30) // 3 = 30 // 3 = 10

# m2 = (0 + 30*2) // 3 = 60 // 3 = 20

m1 = (l*2 + r) // 3

m2 = (l + r*2) // 3

if f(m1, a, b) > f(m2, a, b):

l = m1

else:

r = m2

ans = min(f(l, a, b), f(l+1, a, b), f(r, a, b))

print(f"{ans:.10f}")

メモ

https://www.youtube.com/watch?v=BEkpK0z9yrE&t=2595s

提出

code: python

a, b = map(int, input().split())

# b * i + a / pow(i+1, 0.5)

# for i in range(10):

# print(b * i + a / pow(i+1, 0.5))