I am trying to solve a problem that it want the program to output the result of n^84601. (n=0,1,...,10)
Therefore, I try to solve it by using big integer, and it works well in small number, but segfault in bigger ones.
#include <stdlib.h>
#include <iostream>
using namespace std;
const int MX = 100000;
struct BigInt {
int ar[MX];
int len;
BigInt(int n) {
int i = 0;
while (n != 0) {
ar[i] = n % 10;
n /= 10;
i++;
}
len = i;
}
BigInt times(BigInt x) {
BigInt tmp(0);
for (int i = 0; i < len; i++) {
for (int j = 0; j < x.len; j++) {
int r = ar[i] * x.ar[j] + tmp.ar[i + j];
tmp.ar[i + j] = r % 10;
tmp.ar[i + j + 1] += r / 10;
}
}
for (int i = min(len + x.len, MX - 1);; i--) {
if (tmp.ar[i] != 0) {
tmp.len = i + 1;
break;
}
}
return tmp;
}
void print() {
for (int i = len - 1; i >= 0; i--) {
cout << ar[i];
}
cout << endl;
}
};
BigInt poww(BigInt a, int n) {
if (n == 1) {
return a;
}
BigInt x = poww(a, n / 2);
BigInt y = x.times(x);
if (n % 2 == 1) {
y = y.times(a);
}
return y;
}
int main(void) {
ios::sync_with_stdio(false);
int n;
while (cin >> n) {
if (n == 0)
cout << 0 << endl;
else if (n == 1)
cout << 1 << endl;
else
poww(BigInt(n), 86401).print();
}
return 0;
}
When I change the MX in to 10000 and 86401 into 864, it can correctly caculate 2^864. But it will segfault with 2^86401.
You have a stack overflow.
Your BigInt object is quite large: it contains 100001 ints, which is usually 400,004 bytes.
You allocate several of these on the stack (some are unnecessary: you should really pass arguments by const reference).
You have recursion.
A typical stack size limit is 8MB.
Combine above statements together, and you can see that you can have at most 20 BigInts on the stack at one time. Your recursion depth is at least 17, so creating more than one BigInt on the stack for each recursive call is guaranteed to fail.
There are a few solutions:
use more efficient encoding -- currently you are using int to hold one digit, unsigned char would be more appropriate
allocate space for digits on heap instead of on the stack. If you do that, be aware of the rule of five.
This is what I came up with
#include <iostream>
using namespace std;
int serialNumber = 1;
Would recursion be better?
int factorial(int n)
{
int k=1;
for(int i=1;i<=n;++i)
{
k=k*i;
}
return k;
}
How can I go about doing this in a single for loop?
Or is this the best way?
int main()
{
int a;
int b;
int c;
int fact1;
int fact2;
int fact3;
for (a=1;a < 11;a++)
{
fact1 = factorial(a);
for (b=1;b < 11;b++)
{
fact2 = factorial(b);
for (c=1;c < 11;c++)
{
fact3 = factorial(c);
cout << serialNumber << " : ";
int LHS = fact1 + fact2 + fact3;
if (LHS == a * b * c)
{
cout << "Pass:" <<" "<< a << " & " << b << " & " << c << endl;
}
else
{
cout << "Fail" <<endl;
}
serialNumber++;
}
c = 1;
}
b = 1;
}
return 0;
}
I am being forced to add more none code into it.
Thanks for the help!
Don't know if this is helps,but>
check for minimum of A,B,C
A!+B!+C! = (min(A,B,C)!)*(1+((min+1..restfact1)!)+((min+1..restfact2)!))
So, you can calculate the minimum factorial and than re-use it for calculating others.
On the other hand, you can calculate only the maximum factorial and store its results in the array, and re-use pre-calculated values for finding factorial of smaller numbers
Other implication is that the minimum number can be reduced
restfact1 * restfact2 = ((min-1)!)*(1+((min+1..restfact1)!)+((min+1..restfact2)!))
Part of the question was how can this be done in a single loop and this is one way to do that.
I don't think this is a better way of doing it, but the question was asked:
constexpr int bound = 10;
int Factorials[bound + 1];
for (int i = 1; i <= bound; ++i) Factorials[i] = Factorial(i);
for (int i = 0; i < bound * bound * bound; ++i) {
int s = i + 1;
int a = i;
int c = 1 + a % bound;
a /= bound;
int b = 1 + a % bound;
a /= bound;
++a;
cout << s << " : ";
int LHS = Factorials[a] + Factorials[b] + Factorials[c];
if (LHS == a * b * c)
...
}
In ProjectEuler problem #14, one needs to find the longest Collatz chain, up to 1 million. I found a halfway decent way to do so, however, it feels like I'm just being stupid because I can't find a way to make this code more efficient (the code is supposed to only print out the solution, after it tests 1 to 1 million, but hasn't printed anyting out after 10 minutes). Am I tackling this problem the wrong way, or is there a way to optimize my existing code?
#include <iostream>
using namespace std;
int main()
{
int i;
int x;
int n;
int superN;
int superI;
superN = 0;
superI = 0;
for (i = 1; i <= 1000000; i++) {
x = i;
n = 1;
do {
if (x % 2 == 0) {
x = x / 2;
}
if (x % 2 == 1 && x != 1) {
x = 3 * x + 1;
}
n++;
if (n > superN) {
superN = n;
superI = i;
}
} while (x != 1);
}
cout << "The number " << superI << " ran for " << superN << " terms.";
system("pause");
return 0;
}
You've got a few small problems:
I'm fairly sure that you are overflowing the int data type. Use a uint64_t to make this far less likely.
You should only update superI and superN outside of the while loop. This shouldn't matter, but it hurts performance.
On each iteration you should only modify x once. You currently might modify it twice, which might cause you to fall into an infinite loop. And your calculation of n will be off as well.
Use memoization to improve performance by caching old results.
Applying this, you could come up with some code like this:
#include <cstdint>
#include <iostream>
#include <map>
using namespace std;
int main()
{
uint64_t i;
uint64_t x;
uint64_t n;
uint64_t superN;
uint64_t superI;
std::map<uint64_t, uint64_t> memory;
superN = 0;
superI = 0;
for (i = 1; i <= 1000000; i++) {
x = i;
n = 1;
do {
if (memory.find(x) != memory.end()) {
n += memory[x];
break;
}
if (x % 2 == 0) {
x = x / 2;
} else {
x = 3 * x + 1;
}
n++;
} while (x != 1);
if (n > superN) {
superN = n;
superI = i;
}
memory[i] = n;
}
cout << "The number " << superI << " ran for " << superN << " terms.\n";
system("pause");
return 0;
}
Which takes 4 seconds to output:
The number 837799 ran for 556 terms.
I would suggest not to use memoization as for me it run slower; in my case (up to 10,000,000) the code below is faster without memoization.
the main changes are:
only testing if the current number is even once (not need for an else-if).
using a bitwise operation instead of the modulo operation (slightly faster)
Apart from that I don't know why your code is so long (mine is below 200 milliseconds) maybe you compile as debug ?
bool isEven(uint64_t value)
{
return (!(value & 1));
}
uint64_t solveCollatz(uint64_t start)
{
uint64_t counter = 0;
while (start != 1)
{
if(isEven(start))
{
start /= 2;
}
else
{
start = (3 * start) + 1;
}
counter++;
}
return counter;
}
If you can use compiler intrinsics, particularly with counting and removing trailing zeros, you'll recognize you don't need to branch in the main loop, you'll always alternate odd and even. The memoization techniques that have been previously presented will rarely short-circuit around the math you're doing, since we're dealing with hailstone numbers - additionally, the majority of numbers only have one entry point, so if you see them once, you'll never see them again.
In GCC it'll look something like this:
#include <cstdint>
#include <iostream>
#include <unordered_map>
#include <map>
using namespace std;
using n_iPair = std::pair<uint32_t, uint64_t>;
auto custComp = [](n_iPair a, n_iPair b){
return a.first < b.first;
};
int main()
{
uint64_t x;
uint64_t n;
n_iPair Super = {0,0};
for (auto i = 1; i <= 1000000; i++){
x = i;
n = 0;
if (x % 2 == 0) {
n += __builtin_ctz(x); // account for all evens
x >>= __builtin_ctz(x); // always returns an odd
}
do{ //when we enter we're always working on an odd number
x = 3 * x + 1; // always increases an odd to an even
n += __builtin_ctz(x)+1; // account for both odd and even transfer
x >>= __builtin_ctz(x); // always returns odd
}while (x != 1);
Super = max(Super, {n,i}, custComp);
}
cout << "The number " << Super.second << " ran for " << Super.first << " terms.\n";
return 0;
}
If performance is critical, but memory isn't, you can use caching to improve the speed.
#include <iostream>
#include <chrono>
#include <vector>
#include <sstream>
std::pair<uint32_t, uint32_t> longestCollatz(std::vector<uint64_t> &cache)
{
uint64_t length = 0;
uint64_t number = 0;
for (uint64_t current = 2; current < cache.size(); current++)
{
uint64_t collatz = current;
uint64_t steps = 0;
while (collatz != 1 && collatz >= current)
{
if (collatz % 2)
{
// if a number is odd, then ((collatz * 3) + 1) would result in
// even number, but even number can have even or odd result, so
// we can combine two steps for even number, and increment twice.
collatz = ((collatz * 3) + 1) / 2;
steps += 2;
}
else
{
collatz = collatz / 2;
steps++;
}
}
cache[current] = steps + cache[collatz];
if (cache[current] > length)
{
length = cache[current];
number = current;
}
}
return std::make_pair(number, length);
}
int main()
{
auto start = std::chrono::high_resolution_clock::now();;
uint64_t input = 1000000;
std::vector<uint64_t> cache(input + 1);
auto longest = longestCollatz(cache);
auto end = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::milliseconds>(end - start).count();
std::cout << "Longest Collatz (index : value) --> " << longest.first << " : " << longest.second;
std::cout << "\nExecution time: " << duration << " milliseconds\n";
return EXIT_SUCCESS;
}
I'm trying to get all prime numbers in the range of 2 and the entered value using this c++ code :
#include<iostream>
using namespace std;
int main() {
int num = 0;
int result = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
for (int b = 2; b <= num; b++) {
result = i % b;
if (result == 0) {
result = b;
break;
}
}
cout << result<< endl <<;
}
}
the problem is that I think am getting close to the logic, but those threes and twos keep showing up between the prime numbers. What am I doing wrong?
I've fixed your code and added comments where I did the changes
The key here is to understand that you need to check all the numbers smaller then "i" if one of them dividing "i", if so mark the number as not prime and break (the break is only optimization)
Then print only those who passed the "test" (originally you printed everything)
#include <iostream>
using namespace std;
#include<iostream>
using namespace std;
int main()
{
int num = 0;
int result = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
bool isPrime = true; // Assume the number is prime
for (int b = 2; b < i; b++) { // Run only till "i-1" not "num"
result = i % b;
if (result == 0) {
isPrime = false; // if found some dividor, number nut prime
break;
}
}
if (isPrime) // print only primes
cout << i << endl;
}
}
Many answers have been given which explains how to do it. None have answered the question:
What am I doing wrong?
So I'll give that a try.
#include<iostream>
using namespace std;
int main() {
int num = 0;
int result = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
for (int b = 2; b <= num; b++) { // wrong: use b < i instead of b <= num
result = i % b;
if (result == 0) {
result = b; // wrong: why assign result the value of b?
// just remove this line
break;
}
}
cout << result<< endl <<; // wrong: you need a if-condtion before you print
// if (result != 0) cout << i << endl;
}
}
You have multiple errors in your code.
Simplest algorithm (not the most optimal though) is for checking whether N is prim is just to check whether it doesn't have any dividers in range [2; N-1].
Here is working version:
int main() {
int num = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
bool bIsPrime = true;
for (int b = 2; bIsPrime && b < i; b++) {
if (i % b == 0) {
bIsPrime = false;
}
}
if (bIsPrime) {
cout << i << endl;
}
}
}
I would suggest pulling out the logic of determining whether a number is a prime to a separate function, call the function from main and then create output accordingly.
// Declare the function
bool is_prime(int num);
Then, simplify the for loop to:
for (int i = 2; i <= num; i++) {
if ( is_prime(i) )
{
cout << i << " is a prime.\n";
}
}
And then implement is_prime:
bool is_prime(int num)
{
// If the number is even, return true if the number is 2 else false.
if ( num % 2 == 0 )
{
return (num == 2);
}
int stopAt = (int)sqrt(num);
// Start the number to divide by with 3 and increment it by 2.
for (int b = 3; b <= stopAt; b += 2)
{
// If the given number is divisible by b, it is not a prime
if ( num % b == 0 )
{
return false;
}
}
// The given number is not divisible by any of the numbers up to
// sqrt(num). It is a prime
return true;
}
I can pretty much guess its academic task :)
So here the think for prime numbers there are many methods to "get primes bf number" some are better some worse.
Erosthenes Sieve - is one of them, its pretty simple concept, but quite a bit more efficient in case of big numbers (like few milions), since OopsUser version is correct you can try and see for yourself what version is better
void main() {
int upperBound;
cin >> upperBound;
int upperBoundSquareRoot = (int)sqrt((double)upperBound);
bool *isComposite = new bool[upperBound + 1]; // create table
memset(isComposite, 0, sizeof(bool) * (upperBound + 1)); // set all to 0
for (int m = 2; m <= upperBoundSquareRoot; m++) {
if (!isComposite[m]) { // if not prime
cout << m << " ";
for (int k = m * m; k <= upperBound; k += m) // set all multiplies
isComposite[k] = true;
}
}
for (int m = upperBoundSquareRoot; m <= upperBound; m++) // print results
if (!isComposite[m])
cout << m << " ";
delete [] isComposite; // clean table
}
Small note, tho i took simple implementation code for Sive from here (writing this note so its not illegal, truth be told wanted to show its easy to find)
According to this post, we can get all divisors of a number through the following codes.
for (int i = 1; i <= num; ++i){
if (num % i == 0)
cout << i << endl;
}
For example, the divisors of number 24 are 1 2 3 4 6 8 12 24.
After searching some related posts, I did not find any good solutions. Is there any efficient way to accomplish this?
My solution:
Find all prime factors of the given number through this solution.
Get all possible combinations of those prime factors.
However, it doesn't seem to be a good one.
Factors are paired. 1 and 24, 2 and 12, 3 and 8, 4 and 6.
An improvement of your algorithm could be to iterate to the square root of num instead of all the way to num, and then calculate the paired factors using num / i.
You should really check till square root of num as sqrt(num) * sqrt(num) = num:
Something on these lines:
int square_root = (int) sqrt(num) + 1;
for (int i = 1; i < square_root; i++) {
if (num % i == 0&&i*i!=num)
cout << i << num/i << endl;
if (num % i == 0&&i*i==num)
cout << i << '\n';
}
There is no efficient way in the sense of algorithmic complexity (an algorithm with polynomial complexity) known in science by now. So iterating until the square root as already suggested is mostly as good as you can be.
Mainly because of this, a large part of the currently used cryptography is based on the assumption that it is very time consuming to compute a prime factorization of any given integer.
Here's my code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
#define pii pair<int, int>
#define MAX 46656
#define LMT 216
#define LEN 4830
#define RNG 100032
unsigned base[MAX / 64], segment[RNG / 64], primes[LEN];
#define sq(x) ((x)*(x))
#define mset(x,v) memset(x,v,sizeof(x))
#define chkC(x,n) (x[n>>6]&(1<<((n>>1)&31)))
#define setC(x,n) (x[n>>6]|=(1<<((n>>1)&31)))
// http://zobayer.blogspot.com/2009/09/segmented-sieve.html
void sieve()
{
unsigned i, j, k;
for (i = 3; i<LMT; i += 2)
if (!chkC(base, i))
for (j = i*i, k = i << 1; j<MAX; j += k)
setC(base, j);
primes[0] = 2;
for (i = 3, j = 1; i<MAX; i += 2)
if (!chkC(base, i))
primes[j++] = i;
}
//http://www.geeksforgeeks.org/print-all-prime-factors-of-a-given-number/
vector <pii> factors;
void primeFactors(int num)
{
int expo = 0;
for (int i = 0; primes[i] <= sqrt(num); i++)
{
expo = 0;
int prime = primes[i];
while (num % prime == 0){
expo++;
num = num / prime;
}
if (expo>0)
factors.push_back(make_pair(prime, expo));
}
if ( num >= 2)
factors.push_back(make_pair(num, 1));
}
vector <int> divisors;
void setDivisors(int n, int i) {
int j, x, k;
for (j = i; j<factors.size(); j++) {
x = factors[j].first * n;
for (k = 0; k<factors[j].second; k++) {
divisors.push_back(x);
setDivisors(x, j + 1);
x *= factors[j].first;
}
}
}
int main() {
sieve();
int n, x, i;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> x;
primeFactors(x);
setDivisors(1, 0);
divisors.push_back(1);
sort(divisors.begin(), divisors.end());
cout << divisors.size() << "\n";
for (int j = 0; j < divisors.size(); j++) {
cout << divisors[j] << " ";
}
cout << "\n";
divisors.clear();
factors.clear();
}
}
The first part, sieve() is used to find the prime numbers and put them in primes[] array. Follow the link to find more about that code (bitwise sieve).
The second part primeFactors(x) takes an integer (x) as input and finds out its prime factors and corresponding exponent, and puts them in vector factors[]. For example, primeFactors(12) will populate factors[] in this way:
factors[0].first=2, factors[0].second=2
factors[1].first=3, factors[1].second=1
as 12 = 2^2 * 3^1
The third part setDivisors() recursively calls itself to calculate all the divisors of x, using the vector factors[] and puts them in vector divisors[].
It can calculate divisors of any number which fits in int. Also it is quite fast.
Plenty of good solutions exist for finding all the prime factors of not too large numbers. I just wanted to point out, that once you have them, no computation is required to get all the factors.
if N = p_1^{a}*p_{2}^{b}*p_{3}^{c}.....
Then the number of factors is clearly (a+1)(b+1)(c+1).... since every factor can occur zero up to a times.
e.g. 12 = 2^2*3^1 so it has 3*2 = 6 factors. 1,2,3,4,6,12
======
I originally thought that you just wanted the number of distinct factors. But the same logic applies. You just iterate over the set of numbers corresponding to the possible combinations of exponents.
so int he example above:
00
01
10
11
20
21
gives you the 6 factors.
If you want all divisors to be printed in sorted order
int i;
for(i=1;i*i<n;i++){ /*print all the divisors from 1(inclusive) to
if(n%i==0){ √n (exclusive) */
cout<<i<<" ";
}
}
for( ;i>=1;i--){ /*print all the divisors from √n(inclusive) to
if(n%i==0){ n (inclusive)*/
cout<<(n/i)<<" ";
}
}
If divisors can be printed in any order
for(int j=1;j*j<=n;j++){
if(n%j==0){
cout<<j<<" ";
if(j!=(n/j))
cout<<(n/j)<<" ";
}
}
Both approaches have complexity O(√n)
Here is the Java Implementation of this approach:
public static int countAllFactors(int num)
{
TreeSet<Integer> tree_set = new TreeSet<Integer>();
for (int i = 1; i * i <= num; i+=1)
{
if (num % i == 0)
{
tree_set.add(i);
tree_set.add(num / i);
}
}
System.out.print(tree_set);
return tree_set.size();
}
//Try this,it can find divisors of verrrrrrrrrry big numbers (pretty efficiently :-))
#include<iostream>
#include<cstdio>
#include<cmath>
#include<vector>
#include<conio.h>
using namespace std;
vector<double> D;
void divs(double N);
double mod(double &n1, double &n2);
void push(double N);
void show();
int main()
{
double N;
cout << "\n Enter number: "; cin >> N;
divs(N); // find and push divisors to D
cout << "\n Divisors of "<<N<<": "; show(); // show contents of D (all divisors of N)
_getch(); // used visual studio, if it isn't supported replace it by "getch();"
return(0);
}
void divs(double N)
{
for (double i = 1; i <= sqrt(N); ++i)
{
if (!mod(N, i)) { push(i); if(i*i!=N) push(N / i); }
}
}
double mod(double &n1, double &n2)
{
return(((n1/n2)-floor(n1/n2))*n2);
}
void push(double N)
{
double s = 1, e = D.size(), m = floor((s + e) / 2);
while (s <= e)
{
if (N==D[m-1]) { return; }
else if (N > D[m-1]) { s = m + 1; }
else { e = m - 1; }
m = floor((s + e) / 2);
}
D.insert(D.begin() + m, N);
}
void show()
{
for (double i = 0; i < D.size(); ++i) cout << D[i] << " ";
}
int result_num;
bool flag;
cout << "Number Divisors\n";
for (int number = 1; number <= 35; number++)
{
flag = false;
cout << setw(3) << number << setw(14);
for (int i = 1; i <= number; i++)
{
result_num = number % i;
if (result_num == 0 && flag == true)
{
cout << "," << i;
}
if (result_num == 0 && flag == false)
{
cout << i;
}
flag = true;
}
cout << endl;
}
cout << "Press enter to continue.....";
cin.ignore();
return 0;
}
for (int i = 1; i*i <= num; ++i)
{
if (num % i == 0)
cout << i << endl;
if (num/i!=i)
cout << num/i << endl;
}
for( int i = 1; i * i <= num; i++ )
{
/* upto sqrt is because every divisor after sqrt
is also found when the number is divided by i.
EXAMPLE like if number is 90 when it is divided by 5
then you can also see that 90/5 = 18
where 18 also divides the number.
But when number is a perfect square
then num / i == i therefore only i is the factor
*/
//DIVISORS IN TIME COMPLEXITY sqrt(n)
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
ll int n;
cin >> n;
for(ll i = 2; i <= sqrt(n); i++)
{
if (n%i==0)
{
if (n/i!=i)
cout << i << endl << n/i<< endl;
else
cout << i << endl;
}
}
}
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define MOD 1000000007
#define fo(i,k,n) for(int i=k;i<=n;++i)
#define endl '\n'
ll etf[1000001];
ll spf[1000001];
void sieve(){
ll i,j;
for(i=0;i<=1000000;i++) {etf[i]=i;spf[i]=i;}
for(i=2;i<=1000000;i++){
if(etf[i]==i){
for(j=i;j<=1000000;j+=i){
etf[j]/=i;
etf[j]*=(i-1);
if(spf[j]==j)spf[j]=i;
}
}
}
}
void primefacto(ll n,vector<pair<ll,ll>>& vec){
ll lastprime = 1,k=0;
while(n>1){
if(lastprime!=spf[n])vec.push_back(make_pair(spf[n],0));
vec[vec.size()-1].second++;
lastprime=spf[n];
n/=spf[n];
}
}
void divisors(vector<pair<ll,ll>>& vec,ll idx,vector<ll>& divs,ll num){
if(idx==vec.size()){
divs.push_back(num);
return;
}
for(ll i=0;i<=vec[idx].second;i++){
divisors(vec,idx+1,divs,num*pow(vec[idx].first,i));
}
}
void solve(){
ll n;
cin>>n;
vector<pair<ll,ll>> vec;
primefacto(n,vec);
vector<ll> divs;
divisors(vec,0,divs,1);
for(auto it=divs.begin();it!=divs.end();it++){
cout<<*it<<endl;
}
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
sieve();
ll t;cin>>t;
while(t--) solve();
return 0;
}
We can use modified sieve for getting all the factors for all numbers in range [1, N-1].
for (int i = 1; i < N; i++) {
for (int j = i; j < N; j += i) {
ans[j].push_back(i);
}
}
The time complexity is O(N * log(N)) as the sum of harmonic series 1 + 1/2 + 1/3 + ... + 1/N can be approximated to log(N).
More info about time complexity : https://math.stackexchange.com/a/3367064
P.S : Usually in programming problems, the task will include several queries where each query represents a different number and hence precalculating the divisors for all numbers in a range at once would be beneficial as the lookup takes O(1) time in that case.
java 8 recursive (works on HackerRank). This method includes option to sum and return the factors as an integer.
static class Calculator implements AdvancedArithmetic {
public int divisorSum(int n) {
if (n == 1)
return 1;
Set<Integer> set = new HashSet<>();
return divisorSum( n, set, 1);
}
private int divisorSum(int n, Set<Integer> sum, int start){
if ( start > n/2 )
return 0;
if (n%start == 0)
sum.add(start);
start++;
divisorSum(n, sum, start);
int total = 0;
for(int number: sum)
total+=number;
return total +n;
}
}