Как найти квадраты чисел в диапазоне

You can simplify the arithmetic by using

(n + 1)**2 == n**2 + (2*n + 1)

Here’s how to do that using a generator function:

import math

def squares(lo, hi):
    root = int(math.ceil(lo ** 0.5))
    num = root ** 2
    delta = 2 * root + 1
    while num <= hi:
        yield num
        num += delta
        delta += 2

print list(squares(4, 16))
print list(squares(5, 50))
print list(squares(20, 90))

output

[4, 9, 16]
[9, 16, 25, 36, 49]
[25, 36, 49, 64, 81]

Here’s an equivalent iterator class. I’ve given it a __repr__ method so it looks nice if you print an instance of this class.

import math

class Squares(object):
    def __init__(self, start, stop):
        self.start = start
        self.stop = stop
        root = int(math.ceil(start ** 0.5))
        self.num = root ** 2
        self.delta = 2 * root + 1

    def __repr__(self):
        return 'Squares(%d, %d)' % (self.start, self.stop)

    def __iter__(self): 
        return self

    def next(self):
        num = self.num
        if num > self.stop:
            raise StopIteration
        self.num += self.delta
        self.delta += 2
        return num

sq = Squares(4, 16)
print sq
for i in sq:
    print i

print list(Squares(5, 50))
print list(Squares(20, 90))

output

Squares(4, 16)
4
9
16
[9, 16, 25, 36, 49]
[25, 36, 49, 64, 81]

For Python 3, replace the next method name with __next__.

The usual Python range convention is to stop before you reach the high limit. To make this code comply with that convention, in the squares() generator change

while num <= hi:

to

while num < hi:

and in the Squares() class, change

if num > self.stop:

to

if num >= self.stop:

Вы можете упростить арифметику, используя

(n + 1)**2 == n**2 + (2*n + 1)

Здесь, как это сделать, используя функцию генератора:

import math

def squares(lo, hi):
    root = int(math.ceil(lo ** 0.5))
    num = root ** 2
    delta = 2 * root + 1
    while num <= hi:
        yield num
        num += delta
        delta += 2

print list(squares(4, 16))
print list(squares(5, 50))
print list(squares(20, 90))

Выход

[4, 9, 16]
[9, 16, 25, 36, 49]
[25, 36, 49, 64, 81]

Здесь эквивалентный класс итератора. Я дал ему метод __repr__, поэтому он выглядит хорошо, если вы печатаете экземпляр этого класса.

import math

class Squares(object):
    def __init__(self, start, stop):
        self.start = start
        self.stop = stop
        root = int(math.ceil(start ** 0.5))
        self.num = root ** 2
        self.delta = 2 * root + 1

    def __repr__(self):
        return 'Squares(%d, %d)' % (self.start, self.stop)

    def __iter__(self): 
        return self

    def next(self):
        num = self.num
        if num > self.stop:
            raise StopIteration
        self.num += self.delta
        self.delta += 2
        return num

sq = Squares(4, 16)
print sq
for i in sq:
    print i

print list(Squares(5, 50))
print list(Squares(20, 90))

Выход

Squares(4, 16)
4
9
16
[9, 16, 25, 36, 49]
[25, 36, 49, 64, 81]

Для Python 3 замените имя метода next на __next__.

Обычное соглашение Python range должно остановиться, прежде чем вы достигнете верхнего предела. Чтобы этот код соответствовал этому соглашению, в смене генератора squares()

while num <= hi:

to

while num < hi:

и в классе squares() измените

if num > self.stop:

to

if num >= self.stop:

Вопрос:

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15 ...

1 <= X,Y <= 10^9

0 <= |X-Y| <= 10^6

x=10 , Y=15

ans=5

Лучший ответ:

f(n) = n / 4      => these are divisible by 4, so NOT square-free
       +
       n / 9      => these are divisible by 9, so NOT square-free
       + 
       n / 25     => these are divisible by 16, so NOT square-free
       +
       ...
       -
       n / (4*9)  => these are divisible by both 4 and 9, so we overcounted
       -
       n / (4*25) => these are divisible by both 4 and 25, so we overcounted 
       -
       ...

f(n) = n / (2^2) -
       n / (2^2*3^2) +
       n / (2^2*3^2*5^2) -  <= this becomes 0 very fast. 
                               When it does, 
                               backtrack and increment the previous term.
                               For example, if this were 0,
                               you'd do - n / (2^2*5^2) next
       ...

Ответ №1