用于计数旋转可被 8 整除的 Python3 程序
给定一个大的正数作为字符串,计算给定数的所有可被 8 整除的旋转。
例子:
Input: 8
Output: 1
Input: 40
Output: 1
Rotation: 40 is divisible by 8
04 is not divisible by 8
Input : 13502
Output : 0
No rotation is divisible by 8
Input : 43262488612
Output : 4
方法:对于大数字,很难将每个数字旋转并除以 8。因此,使用“被 8 整除”属性,即如果数字的最后 3 位数字可以被 8 整除,则该数字可以被 8 整除。这里我们实际上并没有旋转数字并检查最后 8 位数字的可分性,而是计算可被 8 整除的 3 位数字的连续序列(以循环方式)。
插图:
Consider a number 928160
Its rotations are 928160, 092816, 609281,
160928, 816092, 281609.
Now form consecutive sequence of 3-digits from
the original number 928160 as mentioned in the
approach.
3-digit: (9, 2, 8), (2, 8, 1), (8, 1, 6),
(1, 6, 0),(6, 0, 9), (0, 9, 2)
We can observe that the 3-digit number formed by
the these sets, i.e., 928, 281, 816, 160, 609, 092,
are present in the last 3 digits of some rotation.
Thus, checking divisibility of these 3-digit numbers
gives the required number of rotations.
Python3
# Python3 program to count all
# rotations divisible by 8
# function to count of all
# rotations divisible by 8
def countRotationsDivBy8(n):
l = len(n)
count = 0
# For single digit number
if (l == 1):
oneDigit = int(n[0])
if (oneDigit % 8 == 0):
return 1
return 0
# For two-digit numbers
# (considering all pairs)
if (l == 2):
# first pair
first = int(n[0]) * 10 + int(n[1])
# second pair
second = int(n[1]) * 10 + int(n[0])
if (first % 8 == 0):
count+=1
if (second % 8 == 0):
count+=1
return count
# considering all
# three-digit sequences
threeDigit=0
for i in range(0,(l - 2)):
threeDigit = (int(n[i]) * 100 +
int(n[i + 1]) * 10 +
int(n[i + 2]))
if (threeDigit % 8 == 0):
count+=1
# Considering the number
# formed by the last digit
# and the first two digits
threeDigit = (int(n[l - 1]) * 100 +
int(n[0]) * 10 +
int(n[1]))
if (threeDigit % 8 == 0):
count+=1
# Considering the number
# formed by the last two
# digits and the first digit
threeDigit = (int(n[l - 2]) * 100 +
int(n[l - 1]) * 10 +
int(n[0]))
if (threeDigit % 8 == 0):
count+=1
# required count
# of rotations
return count
# Driver Code
if __name__=='__main__':
n = "43262488612"
print("Rotations:",countRotationsDivBy8(n))
# This code is contributed by mits.
输出:
Rotations: 4
时间复杂度: O(n),其中n是输入数字的位数。
有关详细信息,请参阅有关可被 8 整除的计数旋转的完整文章!