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.
Examples:
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:
C++
// CPP program to count unary numbers // in a range #include <iostream> using namespace std; // Function to check if a number is unary bool isUnary(int n) { /// Base case. Note that if we repeat // above process for 7, we get 1. if (n == 1 || n == 7) return true; else if (n / 10 == 0) return false; /// rec case int x, sum = 0; while (n != 0) { x = n % 10; sum = sum + x * x; n = n / 10; } isUnary(sum); } // Function to count unary numbers // in a range int countUnary(int a, int b) { int count = 0; for (int i = a; i <= b; i++) { if (isUnary(i) == 1) count++; } return count; } // Driver Code int main() { int a = 1000, b = 1099; cout << countUnary(a, b); return 0; } |
Java
//Java program to count unary numbers // in a range import java.io.*; class GFG { // Function to check if a number is unary static boolean isUnary(int n) { /// Base case. Note that if we repeat // above process for 7, we get 1. if (n == 1 || n == 7) return true; else if (n / 10 == 0) return false; /// rec case int x, sum = 0; while (n != 0) { x = n % 10; sum = sum + x * x; n = n / 10; } return isUnary(sum); } // Function to count unary numbers // in a range static int countUnary(int a, int b) { int count = 0; for (int i = a; i <= b; i++) { if (isUnary(i) == true) count++; } return count; } // Driver Code public static void main (String[] args) { int a = 1000, b = 1099; System.out.println (countUnary(a, b)); } //This code is contributed by ajit } |
Python3
# Python 3 program to count unary numbers
# in a range
# Function to check if a number is unary
def isUnary(n):
# Base case. Note that if we repeat
# above process for 7, we get 1.
if (n == 1 or n == 7):
return True
elif (int(n / 10) == 0):
return False
# rec case
sum = 0
while (n != 0):
x = n % 10
sum = sum + x * x
n = int(n / 10)
return isUnary(sum)
# 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
return count
# Driver Code
if __name__ == ‘__main__’:
a = 1000
b = 1099
print(countUnary(a, b))
# This code is contributed by
# Sanjit_Prasad
C#
//C# program to count unary numbers // in a range using System; public class GFG { // Function to check if a number is unary static bool isUnary(int n) { /// Base case. Note that if we repeat // above process for 7, we get 1. if (n == 1 || n == 7) return true; else if (n / 10 == 0) return false; /// rec case int x, sum = 0; while (n != 0) { x = n % 10; sum = sum + x * x; n = n / 10; } return isUnary(sum); } // Function to count unary numbers // in a range static int countUnary(int a, int b) { int count = 0; for (int i = a; i <= b; i++) { if (isUnary(i) == true) count++; } return count; } // Driver Code public static void Main () { int a = 1000, b = 1099; Console.WriteLine(countUnary(a, b)); } //This code is contributed by 29AjayKumar } |
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