To check a Perfect Number

#include<iostream>
#include<conio.h>
using namespace std;
int main()
{
int number1,number2;
int sum=0;
cout<<"\t\t\t\tPerfect Number";
again:
{

cout<<"\n1. Find in giver range.\n2. Check a number.\nChoice: ";
cin>>number2;
switch (number2)
{
case 1:
{
cout<<"\nEnter lower bound : ";
cin>>number1;
cout<<"Enter upper bound : ";
cin>>number2;
cout<<"Pefect number is  : ";
for(int i=number1;i<=number2;i++)
{
for(int j=1;j<i;j++)
{
if(i%j==0)
{
sum=sum+j;
}
}
if(sum==i)
cout<<i<<"\t";
sum=0;
}
}
break;
case 2:
{
cout<<"Enter Number : ";
cin>>number1;
for(int i=1;i<number1;i++)
if(number1%i==0)
sum=sum+i;
if(sum==number1)
cout<<"Number is perfect";
else
cout<<"Number is not perfect";
}
break;
default:
cout<<"Invalid choice!";
}
cout<<"\nCheck for more number (yes/no) : ";
char ch;
cin>>ch;
if(ch=='y')
goto again;
}
getch();
return 0;
}
Output: -

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