编写函数,把一个自然数分解成最多4个平方数的和,然后调用这个函数。例如7604=2704+4900=52²+70²。
import itertools
n=eval(input('enter a number:')) #输入一个数
a=range(1,int(n**0.5)+1) #构造一组数,从1到n的开方,作为备选答案
haveAnswer=False #已经找到解
for i in range(2,5): # 循环,i表示从a里取出几个数来
b=itertools.combinations_with_replacement(a,r=i) #有放回的组合(可重复选择)
for b1 in b:
c=sum([x**2 for x in b1]) #求平方和
if c==n:
haveAnswer=True
print('{}={}'.format(n,' * '.join([str(x)+'^2' for x in b1])))
break
if haveAnswer:
break
给个例子参考:
from math import sqrt
def fun(n):
for a in range(int(sqrt(n))):
for b in range(int(sqrt(n))):
if a**2+b**2>n:
break
for c in range(int(sqrt(n))):
if a**2+b**2+c**2>n:
break
for d in range(int(sqrt(n))):
if a**2+b**2+c**2+d**2==n:
return [a,b,c,d]
n = 7604
result = fun(n)
lst = [r for r in result if r>0]
print(lst)
def cal(n):
for i in range(n+1):
for j in range(n+1):
for k in range(n+1):
for l in range(n+1):
if i*i+j*j+k*k+l*l == n:
return i,j,k,l
n = int(input())
a,b,c,d=cal(n)
print("{}={}+{}+{}+{}={}+{}+{}+{}".format(n,a**2,b**2,c**2,d**2,a,b,c,d))
from itertools import combinations
def sqrtsum(num):
res = []
lst = [i**2 for i in range(1, int(num**0.5)+1)]
for i in range(1,5):
for j in combinations(lst,i):
if sum(j) == num:
for k in j:
res.append(int(k**0.5))
return res
return res