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Program Based on Loops

Program to check whether a number is perfect or not

Input - Output

Input
Enter a number : 6

Output
6 is a perfect number

Input
Enter a number : 8

Output
8 is not a perfect number

Algorithm

Step 1: START

Step 2: Enter a number 'num' as input.

store value of num in temp variable and sum=0;

Step 3: Start a loop: for(int i=1;i< num;++i)

  if num % i==0;

  sum=sum+i;

Step 4:end of loop

Step 5: if sum==temp

     Print It is a perfect number

Else Print It is not a perfect number

step 6: End

CODE


/* Check whether a number is perfect or not */

#include<stdio.h>
int main()
{
      int num,sum=0;
      printf(" Enter an integer number : ");
      scanf("%d",&num);
      for(int i=1;i<num;i++)
    {
       if(num % i == 0)
        sum += i;
    }
      if(sum == num)
       printf("%d is a perfect number",num);
      else
       printf("%d is not a perfect number",num);
    return 0;
}


/* Check whether a number is perfect or not */

#include<iostream>
using namespace std;
int main()
{
      int num,sum=0;
      cout<<" Enter an integer number : ";
      cin>>num;
      for(int i=1;i<num;i++)
    {
       if(num % i == 0)
        sum += i;
    }
      if(sum == num)
       cout<<" "<<num<<" is a perfect number";
      else
       cout<<" "<<num<<" is not a perfect number";
    return 0;
}

OutPut

Enter a number : 6

6 is a perfect number

Recommendation

1. Basic C program.
2. Program based on If - Else
3. Program based on Loops
4. Program based on Conversion
5. Program based on Pattern