## Problem Statement

We need to write a program, that accepts a positive integer value from the user and print all the possible divisors of that number (excluding the number itself).

**For example**

Input:100Output :1 2 4 5 10 25 50

**Algorithm**

The problem statement is quite straightforward and beginner-level. To solve this problem all we need is a loop and a conditional `if`

statement.

As the problem statement state that all we need to do is find all the divisors except for the number itself. So we need a loop from range 1 to half of the number entered by the user. Because a number’s all possible divisors can be the number itself and less than half of it.

### C program to find the divisors of a Positive Integer

#include <stdio.h> int main() { int num, i; //input the number printf("Enter a positive integer value: "); scanf("%d",&num); for(i=1; i<=num/2; i++) { if(num%i==0) { printf("%d ", i); } } return 0; }

**Output**

Enter a positive integer value: 100 1 2 4 5 10 20 25 50

**C++ **program to find the divisors of a Positive Integer

#include<iostream> using namespace std; int main() { int num; //input the positive integer cout<<"Enter a Positive Integer Number: "; cin>>num; for(int i=1; i<=num/2; i++) { if(num%i==0) { cout<<i<<" "; } } return 0; }

**Output:**

Enter a Positive Integer Number: 100 1 2 4 5 10 20 25 50

**Python:**

# input number num = int(input("Enter a Positive Integer Number: ")) for i in range(1, (num//2)+1): if num%i==0: print(i, end =" ")

**Output:**

Enter a Positive Integer Number: 100 1 2 4 5 10 20 25 50

### Wrapping Up!

Now let’s wrap up this programming tutorial on “Write a program to find the divisors of a positive integer”. To find all the divisors of a number `n`

we just need to calculate all smaller numbers that can divide the number `n`

completely. If we look closely all the divisor of a number lies between 1 to half of the number itself, which save the time and we just need to create a loop that would iterate n/2 time.

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