Python 把一个数分成两个幂的和
x是我的输入。 我需要找到I,j>=0和n,m>1,比如Python 把一个数分成两个幂的和,python,algorithm,math,optimization,Python,Algorithm,Math,Optimization,x是我的输入。 我需要找到I,j>=0和n,m>1,比如x=I**m+j**n 现在我一直在这样做,但这是一个缓慢的方式!!我怎样才能改进它 from math import sqrt import numpy as np def check(x): for i in range(1,int(np.ceil(sqrt(x)))): for j in range(1,int(np.ceil(sqrt(x)))): for m in range(2,
x=I**m+j**n
现在我一直在这样做,但这是一个缓慢的方式!!我怎样才能改进它
from math import sqrt
import numpy as np
def check(x):
for i in range(1,int(np.ceil(sqrt(x)))):
for j in range(1,int(np.ceil(sqrt(x)))):
for m in range(2,x/2+1):
for n in range(2,x/2+1):
if((pow(i,m) +pow(j,n))==x):
print 'Yes';
return ;
print 'No';
谢谢大家! 这里不时会出现一个问题,关于确定一个正整数是否是另一个正整数的整数幂。即,给定正整数
z
查找正整数j
和n
,从而z==j**n
。这可以在时间复杂度O(log(z))
中完成,因此速度相当快
因此,找到或开发这样一个例程:将其称为is_a_power(z)
,如果z
是(0,0)
的幂,则返回一个元组(j,n)
。然后在i
和m
上循环,然后检查x-i**m
是否为电源。当它成为一个,你就完成了
我将让您从这里完成代码,除了一个指针。给定
x
和i
其中i>1
,您可以找到m
的上限,这样i**m您就可以通过找到所有小于x的幂(i**m)来反转过程。然后你只需检查这些幂的任何一对加起来是否等于x
def check(x):
all_powers = set([1]) #add 1 as a special case
#find all powers smaller than x
for base in range(2,int(math.ceil(sqrt(x)))):
exponent = 2;
while pow(base, exponent) < x:
all_powers.add(pow(base, exponent))
exponent+=1
#check if a pair of elements in all_powers adds up to x
for power in all_powers:
if (x - power) in all_powers:
print 'Yes'
return
print 'No'
def检查(x):
所有_powers=set([1])#添加1作为特例
#查找所有小于x的幂
对于范围内的基(2,int(math.ceil(sqrt(x))):
指数=2;
当pow(基准,指数)
上面的代码很简单,但可以进行优化,例如,通过集成while循环中检查一对是否等于x,在大多数情况下,您可以提前停止。来自math import sqrt
from math import sqrt
import numpy as np
def build(x):
# this function creates number that are in form of
# a^b such that a^b <= x and b>1
sq=sqrt(x);
dict[1]:1; # 1 is always obtainable
dict[0]:1; # also 0 is always obtainable
for i in range(1,sq): # try the base
number=i*i; # firstly our number is i^2
while number<=x:
dict[number]:1; # this number is in form of a^b
number*=i; # increase power of the number
def check(x):
sq=sqrt(x);
for i in range(1,sq): # we will try base of the first number
firstnumber=1;
while firstnumber<=x: # we are trying powers of i
remaining=x-firstnumber; # this number is remaining number when we substract firstnumber from x
if dict[remaining]==1: # if remaining number is in dictionary which means it is representable as a^b
print("YES"); # then print YES
return ;
firstnumber*=i; # increase the power of the base
print("NO");
return ;
将numpy作为np导入
def生成(x):
#此函数用于创建以下形式的数字:
#a^b使a^b 1
sq=sqrt(x);
dict[1]:1;#1总是可以获得的
dict[0]:1;#而且0总是可以获得的
对于范围(1,sq)内的i:#尝试底部
数字=i*i;#首先,我们的号码是i^2
虽然数字这可能不是问这个问题的最佳地方,但你可以做的一件事是,如果值总是超过x,请确保退出循环:如果x是10,并且你有i=3
,那么在m>3
时检查任何内容都是毫无意义的,因为它总是大于10,我可以详细说明一下,但请告诉我,如果这是您想要的方向,那么执行int(np.ceil(x))
的目的是什么?存储x/2+1
的结果并在m和n次迭代中使用它可能会更快?
from math import sqrt
import numpy as np
def build(x):
# this function creates number that are in form of
# a^b such that a^b <= x and b>1
sq=sqrt(x);
dict[1]:1; # 1 is always obtainable
dict[0]:1; # also 0 is always obtainable
for i in range(1,sq): # try the base
number=i*i; # firstly our number is i^2
while number<=x:
dict[number]:1; # this number is in form of a^b
number*=i; # increase power of the number
def check(x):
sq=sqrt(x);
for i in range(1,sq): # we will try base of the first number
firstnumber=1;
while firstnumber<=x: # we are trying powers of i
remaining=x-firstnumber; # this number is remaining number when we substract firstnumber from x
if dict[remaining]==1: # if remaining number is in dictionary which means it is representable as a^b
print("YES"); # then print YES
return ;
firstnumber*=i; # increase the power of the base
print("NO");
return ;