CareerCup

The loop can be optimized:
for(int i = 1; i <= n/2; i++)

The number is nothing but a perfect number. Here is the java code to find the perfect number:

public void isPerfectNumber(int n)
{
int sum = 0;

for(int i = 1; i <= n-1; i++)
{
if(n%i == 0)
{
sum = sum + i;
}
}
if(sum == n)
System.out.println("The number is a perfect number");
else
System.out.println("The number is not a perfect number");
}

Python code:
def sumOfFactor(N):
oN = N
F = 2
SUM = 0;
while F < oN :
if N % F == 0:
SUM += F
N = N / F
#print(N)
else:
F+=1
return SUM == oN

for i in range(10000): #test 0 - 10000
if sumOfFactor(i):
print i

seems only 4 is satisfied?

bool isPerfect(int n)
{

  int sum=0;
  int m=n;		//save the original value
  
  int i=2;
  while (i<n)
  {
     if ( 0 == n%i)
     { 
        sum += i;
        n = n/i         // keep on another factor
        i=2;		//restart from 2
     }
     i++;
  }

 if (m!=n))sum += n;   //count the last factor

 return (sum == m);
}

The answer written by maggie is correct

How about following solution?

Idea is to take scan from 2 to sqrt(n) and add i and n/i if n%i == 0;
eg.


for n = 12,
sqrt(n) ~ 3
sum = 1 + (2 + 6) + (3 + 4)

bool isPerfectNumber(int n)
{
int sum = 0;
int mid = sqrt(n);

if(mid*mid == n) sum+=mid;
if(n>0) sum+=1;

for(int i = 2 ; i < mid;i++)
{
if(n%i == 0) sum += i + n/i;
}

return sum == n;
}

yaa. i meant first. sorry for the typo.
i expected a comment on the algo :P

public class General {

int add = 0;
static int originalValue = 100;
public static void main(String[] args) {
if(args[0] != null){
originalValue = Integer.parseInt(args[0]);
}

General gn = new General();
gn.findFactor(originalValue, 2);
}

public int findFactor(int number,int divisor){
int value = number%divisor;
int returnvalue = -1;
if(value == 0){
returnvalue = number/divisor;
System.out.println(divisor);
add = add + divisor;
if(returnvalue == 1){
if(divisor == originalValue){
System.out.println("No factors, prime number");
}
else{
if(add == originalValue){
System.out.println("You are lucky, factors add to original value");
}
else{
System.out.println("No, the factors do not add to original value");
}

}
}else{
findFactor(returnvalue,divisor);
}
}
else{
findFactor(number,divisor+1);
}
return returnvalue;
}
}

//============================================================
//Program to check if a number is sum of its factors or not
#include <iostream>
using namespace std;

int sumofFactors(int n) {
    int i=2;
    int sum = 0;
    int original = n;
    while(n!=1) {
        if(n%i==0) {
            n = n/i;
            sum += i;
            i = 2;
        }
        else i++;
    }
    return (sum == original);
}

int main(int argc,char **argv) {
    int number;
    cout<<"Enter a number: ";
    cin>>number;
    cout<<sumofFactors(number)?"TRUE":"FALSE";
}
//============================================================

it is not a perfect square, perfect square is the one that has odd number of factors but sum of factors not equal to number for ex: 25 factors are 1, 5, sum = 6. but solution by maggie is perfect except that it is not a perfect square.

//java code
boolean checksum(int x){
if (x <= 1)
return false;
int y = 2;
int count = 0;
while (y * Y <= x){
if (x % y == 0)
count += x + (x/y);
x++;
}

return (count + 1 == x;
}

<pre lang="" line="1" title="CodeMonkey38712" class="run-this">// Perfect number
// Using trial Division technique

#include <iostream>
#include <cstdio>
#include <vector>

using namespace std;

int main(){
int n;
cin>>n;
vector<int> factors;
factors.push_back(1);
for(int i=2;i*i<=n;i++){
if(n%i==0){
factors.push_back(i);
factors.push_back(n/i);
}
}
int result=0;
for(int i=0;i<factors.size();i++){
result+=factors[i];
}
if(result==n)
cout<<"Perfect Number"<<endl;
else
cout<<"Not a perfect number"<<endl;
return 0;
}</pre>

Is the "1" counted to the factors of one number?

<pre lang="" line="1" title="CodeMonkey65333" class="run-this">public void sumFactor (int num) {

int sum = 0;
int max = (int) num/2;

for (int i = 1; i <= max; i++) {
if (num%i==0)
sum += i;
}
if (sum == num) {
System.out.println (num + " is the sum of its factors");
}
}

</pre><pre title="CodeMonkey65333" input="yes">
</pre>

#include<stdio.h>
#include<iostream>
#include<math.h>

int main()
{
    int no=0,i=0;
    printf("Enter number\n");
    scanf("%d",&no);
    int sum=1;
    int psuedoNo=no;
    for(i=2;i<=sqrt(no);i++)
    {
                            if(no%i==0)
                            {
                                             sum=sum+i+no/i;
                            }
                            
    }              
        
        if(sum==no)
        printf("PErfect number");
        else
        printf("Not PErfect number");
        system("PAUSE");
}

Complexity O(sqrt(n))

#include<stdio.h>
main()
{ int i,j,n,tmp=0,a[30]={0},sum=0;
printf("Enter the number upto which you want output\n");
scanf("%d",&n);
for(i=2; i<=n ; i++)
{ tmp=0;
sum=0;
for(j=1; j<=i/2+1; j++)
{ if(i%j==0)
{ a[tmp]=j;
tmp++;
sum=sum+j;
}

}
if(sum==i)
printf("%d, ",sum);
}
}

public class factor {

	public static void main(String args[]) {
		int number = 496;
		int sum = 0;
		int i = 1;
		while (i <= number / 2) {
			if (number % i == 0) {
				sum = sum + i;
			}
			i++;
		}
		if (number == sum)
			System.out.println("perfect");
		else
			System.out.println("not perfect");
	}
}

public boolean isPerfectNumber(int n)
	{
		int i=0, sum=1;
		for(i=2;i*i<n;i++)
		{
				if(n%i==0)
				{
					sum+=i;
					sum += (n/i);
				}
		}
		if(i*i==n)
			sum+=i;
		return (sum==n);
	}

public class sumFactorNumber {

		public static void main(String[] args){
			
			for(int n = 2; n<=100; n++)
			{
			int sum = 1; //1 is the factor of all numbers  
			
			
			for(int i=2; i<n; i++)
			{
				if(n%i==0)
				{
					sum = sum + i;
				}
			}
			
			
			if(sum==n)
				System.out.println(n+"is sum factor");
			}
		}
}