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; ` `} ` |

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## 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 ` `} ` |

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## 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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**Output:**

13

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