Python：每日一题 3

892427369

for x in range(100000):
    if (x+100)**0.5%1 == 0 and (x+268)**0.5%1 == 0:
        print(x)

DavidCowboy

count = 100000
while (count + 100) ** 0.5 != int((count + 100) ** 0.5) or (count + 268) ** 0.5 != int((count + 268) ** 0.5):
    count -= 1
print(count)

uilUVUBWP

import math

for i in range(1000):
    a = math.sqrt(i+100)
    b = math.sqrt(i+268)
    if a*a == (i+100) and b*b == (i +268):
        print(i)

songmenghua

import math
for i in range(10000):
    a = (i+100)**0.5
    b = (i+268)**0.5
    if int(a)==a and int(b)==b:
        print '该数为：', i

d2nte

强行骚……

import math
import itertools
numbers = itertools.count(-100)
for number in numbers:
    a = math.sqrt(number + 100)
    b = math.sqrt(number + 268)
    if a == int(a) and b == int(b):
        print(number)

acgods

1. from math import *
   
2. list1 = []
   
3. for number in range(0,10000):
   
4.     x = int(sqrt(number + 100))
   
5.     y = int(sqrt(number + 268))
   
6.     if (number + 100 == pow(x,2)) and (number + 268 == pow(y,2)):
   
7.         list1.append(number)
   
8. print(list1)

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修改后的代码

1. list1 = []
   
2. for number in range(0,10000):
   
3.     if(sqrt(number + 100)%1 ==0) and (sqrt(number + 268)%1 ==0):   
   
4.         list1.append(number)
   
5. print(list1)

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爱城

新手·ing 发表于 2017-3-25 10:00
我的解答！

import math
for i in range(10000):
    a=math.sqrt(i+100)
    b=math.sqrt(i+268)
    c=int(a)
    d=int(b)
    if a*a==c*c and b*b==d*d:
        print(i)



较之答案，有些冗余

god圣锋

1. import math
   
2. 
   
3. count = 0
   
4. 
   
5. for i in range(-100,10001):
   
6. 
   
7.     if math.sqrt(i+100)% 1 == 0.0 and math.sqrt(i+268)% 1 == 0.0:
   
8. 
   
9.         print(i,'\t',end = ' ')
   
10. 
    
11.         count += 1
    
12.         
    
13.         if not count %5:
    
14.             print('\n')
    
15. 
    
16. print('\n\n总共有%d个数\n' % count)

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程序员的救赎

冬雪雪冬 发表于 2017-3-25 10:33
我也写一个

问个问题，把len(list1) 写在前面用个变量代替的话，会不会更节约时间？即每次循环会不会重复计算len(list1)？

sxzpf

import math
for i in range(-100,10001):
    if math.sqrt(i+100)%1==0 and math.sqrt(i+268)%1==0:
        print(i)

sxzpf

import math
for i in range(-100,10005,1):
    a = int(math.sqrt(i+100))
    b = int(math.sqrt(i+268))
    if a * a == (i+100) and b * b == (i+268):
        print(i)

_玛莎_

import math
for i in range(10000):
    x = int(math.sqrt(i+100))
    y = int(math.sqrt(i+268))
    if x * x == (i+100) and y * y == (i+268):
        print(i)

_玛莎_

import math
for i in range(10000):
    x = int(math.sqrt(i+100))
    y = int(math.sqrt(i+268))
    if x * x == (i+100) and y * y == (i+268):
        print(i)

liujian973

    number=1
    import math
    while True:
        if not math.sqrt(number+100)%1 and not math.sqrt(number+268)%1:
            print(number)
            if number>10000:break
        number +=1

Roc乘风

一行的那位大神，思路真六

Roc乘风

pizzapasta 发表于 2017-4-11 08:28
import math

for x in range(10000):

请教  ，这个is_integer()  是用来判断  一个数字是否是整数吗？

百度没有搜到详细的 相关

zhangjk19841984

import math
for a in range(100000):
    b=int(math.sqrt(a+100))
    c=int(math.sqrt(a+268))
    if b*b==a+100 and c*c==a+268:
        print(a)

葑纆

1. import math
   
2. 
   
3. for x in range(10001):
   
4.     y1 = math.sqrt(x + 100)
   
5.     y2 = math.sqrt(x + 268)
   
6.     if y1 % int(y1) == 0 and y2 % int(y2) == 0:
   
7.         print(x)

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I_love_fishc_tj

from math import sqrt
def number():
    for i in range(-100,10000):
        if sqrt(i + 101)%1 == 0 and sqrt(i + 269)%1 == 0:
            print(i)

I_love_fishc_tj

solomonxian 发表于 2017-4-19 18:38
整数%1得0， 小数会带尾巴

看到这个余1，才知道自己编写的时候，判断条件少了什么~  感谢。