修改后的Kaprekar数是带有d位的正整数n,因此当我们将其平方分为两部分时-右手r为d位,左手为l包含剩余的d或d-1位,总和等于原始数(即l + r = n)。
您可以通过测试给定范围内给定条件的每个数字来找到给定范围内的Kaprekar数。
def print_Kaprekar_nums(start, end): for i in range(start, end + 1): # Get the digits from the square in a list: sqr = i ** 2 digits = str(sqr) # Now loop from 1 to length of the number - 1, sum both sides and check length = len(digits) for x in range(1, length): left = int("".join(digits[:x])) right = int("".join(digits[x:])) if (left + right) == i: print("Number: " + str(i) + "Left: " + str(left) + " Right: " + str(right)) print_Kaprekar_nums(150, 8000)
输出结果
这将给出输出-
Number: 297Left: 88 Right: 209 Number: 703Left: 494 Right: 209 Number: 999Left: 998 Right: 1 Number: 1000Left: 1000 Right: 0 Number: 2223Left: 494 Right: 1729 Number: 2728Left: 744 Right: 1984 Number: 4879Left: 238 Right: 4641 Number: 4950Left: 2450 Right: 2500 Number: 5050Left: 2550 Right: 2500 Number: 5292Left: 28 Right: 5264 Number: 7272Left: 5288 Right: 1984 Number: 7777Left: 6048 Right: 1729