problem 21

/*
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 284 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(284) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 284.

Evaluate the sum of all the amicable numbers under 10000
*/
#include<iostream>
#include<math.h>
#include<map>
#include<vector>
#include<time.h>       
using namespace std;

  time_t timer;


int sumofdiv(int n){
   int sum = 0;
  for( int i = 1; i <= n / 2 ; i++){
    if( n % i == 0){
      sum = sum + i;
    }
  }
  return sum;
}

int main(){
  int sumof = 0;
  for(int i = 2; i< 10000; i++){
     int sum1 = sumofdiv(i);
     if(sumofdiv(sum1) == i && i!= sum1){
       sumof = sumof + i + sum1;
     }
  }
 cout<<sumof / 2<<endl;

  return 0;
}