for j in range(1, i):
if sum(lst) == i:
def lazy_mul(n):
def mul(m):
return m*n
return mul
需求分析:
解决方案:
编写一个函数,该函数可以接收一个整数N作为参数,然后返回小于等于N的所有完美数和它们的因子。
首先需要判断N是否小于等于0,若小于等于0直接返回空结果。
python def find_perfect_numbers(N): if N <= 0: return []
```python def find_perfect_numbers(N): if N <= 0: return []
perfect_numbers = []
factors = {}
for num in range(2, N+1):
sum_factors = sum([i for i in range(1, num) if num%i==0])
if sum_factors == num:
perfect_numbers.append(num)
factors[num] = [i for i in range(1, num) if num%i==0]
return perfect_numbers, factors
```
python def print_results(perfect_numbers, factors): for num in perfect_numbers: print(f"{num}: {factors[num]}")
python N = int(input("请输入一个整数N: ")) perfect_numbers, factors = find_perfect_numbers(N) print_results(perfect_numbers, factors)
完整代码如下:
def find_perfect_numbers(N):
if N <= 0:
return []
perfect_numbers = []
factors = {}
for num in range(2, N+1):
sum_factors = sum([i for i in range(1, num) if num%i==0])
if sum_factors == num:
perfect_numbers.append(num)
factors[num] = [i for i in range(1, num) if num%i==0]
return perfect_numbers, factors
def print_results(perfect_numbers, factors):
for num in perfect_numbers:
print(f"{num}: {factors[num]}")
N = int(input("请输入一个整数N: "))
perfect_numbers, factors = find_perfect_numbers(N)
print_results(perfect_numbers, factors)