Solve this algorithm

//shadow gives ghost a number n
//if n =6 ghost has to give 3 cookies
//if n =2 ghost has to give 1 cookies
//work for splitting in two parts
//it will always break number into two parts
//and the no.of cookie will always 2

#include <iostream>

using namespace std;

typedef struct set{
int x;
int y;
}vector;

vector dataset[100];
int count = 1;

int fact(int n){
int j=1;


for(int i=1;i<n;i++){
if(n%i == 0){
j=i;
}
}
return j;
}

//we have to introduce prime-number concept so that we can minimize the no.of cookies
bool isPrime(int n)
{

if (n <= 1) return false;


for (int i=2; i<n; i++)
if (n%i == 0)
return false;

return true;
}

int split(int n){

int m=0;
for(int i=2;i<n;i++){
for(int j=2;j<n;j++){
if(i+j == n){
count++;
dataset[m].x = i;
dataset[m].y = j;
m = m+1;
}
}
}
}

int findMin(int arr[],int n){
int minthisfunc = arr[0];
int i ;
int minindex = 0;
for( i=1;i<n;i++){
if(minthisfunc > arr[i]){
minthisfunc = arr[i];
}
}

for( i=1;i<n;i++){
if(minthisfunc == arr[i]){
return i;
}
}

}

int main()
{
int n;
cout<<"give number"<<endl;
cin>>n;

if(n>0 && n<10){
if(isPrime(n)){
cout<<n<<endl;
return 0;
}
}

split(n);

int min[1000] ;
int j=0;




for(int i=0;i<count-1;i++){
min[j] = fact(dataset[i].x)+fact(dataset[i].y);
++j;
}

int ans = findMin(min,count-1);

cout<<"no.of cookies->"<<fact(dataset[ans].x)+fact(dataset[ans].y)<<endl;
cout<<"("<<dataset[ans].x<<","<<dataset[ans].y<<")";

return 0;
}