##
**
Problem Statement
**

We need to write a program that accepts a positive integer value from the user and prints 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 time, and we just need to create a loop that would iterate n/2 time.

**
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