CareerCup

am I missing something in this question ?
all you need to find is , if given bar nTiles can be divided by width. if it can be and remainder is less than width , then one break is required.
Otherwise, find two factors , if two integer factors are possible then 2 break are required.

I have solved with code and alog : Please check gohired dot in/2014/07/rectangular-chocolate-bar-create-at dot html

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <cstdlib>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
while(true)
{


long long x,y,n;
cin>>x>>y>>n;
if(x*y<n)
{
cout<<"-1"<<endl;
}
else if(x*y==n)
{
cout<<"0"<<endl;
}
else if((n%x==0 && (n/x)<y) || (n%y==0 && (n/y)<x))
{
cout<<"1"<<endl;
}
else
{
bool f=false;
for(int i=1;i<=sqrt(n);i++)
{
if(n%i==0)
{
int a=i;
int b=n/i;
if((a<=x && b<=y) || (a<=y && b<=x))
{
f=true;
cout<<"2"<<endl;
break;
}
}
if(f==true)
{
break;
}
}
if(f==false)
{
cout<<"-1"<<endl;
}
}
}

return 0;
}

int getMinSplit(int height, int width, int nTiles) {
	if(nTiles>height*width) return -1;
	else if(nTiles==height || nTiles==width) return 1;
	else if(nTiles==height*width) return 0;
	else return 2;
}