我有一个生成三角数的代码,计算每个三角数的除数,并打印出3列,其中第一列是输入数,第二列是三角数,第三列是每个三角形的除数数。如何计算每个三角数的除数(计算第三列中的数字)?
for num in range(10):
triangle_number = num*(num+1)//2
print(num, end = ' ')
print(triangle_number, end = ' ')
for divisor in range(1, triangle_number+1):
if triangle_number%divisor == 0:
print(divisor, end = ',')
print()
输出看起来像这样:
0 0
1 1 1,
2 3 1,3,
3 6 1,2,3,6,
4 10 1,2,5,10,
5 15 1,3,5,15,
6 21 1,3,7,21,
7 28 1,2,4,7,14,28,
8 36 1,2,3,4,6,9,12,18,36,
9 45 1,3,5,9,15,45,
答案 0 :(得分:0)
您可以在for
循环中添加一个变量,并在获得除数并打印结尾时对其进行递增。
for num in range(10):
count = 0
triangle_number = num*(num+1)//2
print(num, end = '\t')
print(triangle_number, end = '\t')
for divisor in range(1, triangle_number+1):
if triangle_number%divisor == 0:
print(divisor, end = ',')
count += 1
print("\t", count)
##Output:
##0 0 0
##1 1 1, 1
##2 3 1,3, 2
##3 6 1,2,3,6, 4
##4 10 1,2,5,10, 4
##5 15 1,3,5,15, 4
##6 21 1,3,7,21, 4
##7 28 1,2,4,7,14,28, 6
##8 36 1,2,3,4,6,9,12,18,36, 9
##9 45 1,3,5,9,15,45, 6
##Output order : input number, count of divisors divisors, triangular number
或者您可以在打印前创建除数列表并测量长度。
for num in range(10):
divisors = []
triangle_number = num*(num+1)//2
print(num, end = '\t')
print(triangle_number, end = '\t')
for divisor in range(1, triangle_number+1):
if triangle_number%divisor == 0:
divisors.append(divisor)
print(len(divisors), "\t", ",".join(str(div )for div in divisors))
##Output:
##0 0 0
##1 1 1 1
##2 3 2 1,3
##3 6 4 1,2,3,6
##4 10 4 1,2,5,10
##5 15 4 1,3,5,15
##6 21 4 1,3,7,21
##7 28 6 1,2,4,7,14,28
##8 36 9 1,2,3,4,6,9,12,18,36
##9 45 6 1,3,5,9,15,45
##Output order : input number, triangular number, count of divisors divisors
答案 1 :(得分:0)
您可以使用列表理解替换您的内部for循环,然后使用CoroutineScope
函数获取该列表的长度,即替换:
len
具有:
for divisor in range(1, triangle_number+1):
if triangle_number%divisor == 0:
print(divisor, end = ',')
列表理解是创建列表的一种简洁方法,可以通过在divisors = [divisor for divisor in range(1, triangle_number+1) if triangle_number%divisor == 0]
print(*divisors, sep=',', end=' ')
print(len(divisors), end='')
内append
进行创建,如果您拥有列表{{1},for loop
(星号运算符)会解压缩},那么*
等同于x = [1, 2, 3]
,依此类推。