Given two numbers A and B, A<=B, the task is to find the number of unary numbers between A and B, both inclusive.
Unary Number: Consider the number 28. If we take the sum of square of its digits, 2*2 + 8*8, we get 68. Taking the sum of squares of digits again, we get 6*6 + 8*8=100. Doing this again, we get 1*1 + 0*0 + 0*0 = 1. Any such number, which ultimately leads to 1, is called a unary number.
Input : A = 1, B = 10 Output : 3 Input : A = 100, B = 150 Output : 7
The idea is to recursively calculate sum of squares of digits of the number and every time recurring down replace the number with calculated sum.
The base cases of the recursion will be:
- If the sum if reduced to either 1 or 7, then answer is true.
- If the sum if reduced to a single digit integer other than 1 and 7, answer is false.
Below is the recursive implementation of this problem:
# Python 3 program to count unary numbers
# in a range
# Function to check if a number is unary
# Base case. Note that if we repeat
# above process for 7, we get 1.
if (n == 1 or n == 7):
elif (int(n / 10) == 0):
# rec case
sum = 0
while (n != 0):
x = n % 10
sum = sum + x * x
n = int(n / 10)
# Function to count unary numbers
# in a range
def countUnary(a, b):
count = 0
for i in range(a, b + 1, 1):
if (isUnary(i) == 1):
count += 1
# Driver Code
if __name__ == ‘__main__’:
a = 1000
b = 1099
# This code is contributed by
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