CUTTING RECIPES

c-plus-plus
easy
recipe

#1

Problem: http://www.codechef.com/problems/RECIPE
difficulty:easy
Hello,
I have tried many times but i am unable to point out any errors (logical) errors in my code.I am getting a wrong answer for this code.It works fine on my machine giving correct output.Can anybody help me out.
I have found the two smallest numbers(min1 and min2) and then their gcd.
If the gcd is not 1 then i am checking whether their gcd divides all the remaining nos without leaving behind any remainder.If it is 1 i simply print all nos.
If so then i divide all the numbers by the gcd and output the result.
Else i simply output all the numbers.

#include<iostream>

using namespace std;

int main()
{
int t,n,i,min1,min2,arr[50];
cin>>t;
while(t–)
{

    cin>>n;
    i=0;
    //since max value of any no. can be 1000
    min1=min2=1001;
    //min1<min2
    while(i<n)
    {
        cin>>arr*;
        if(arr*<min1)
        {
             min2=min1;
             min1=arr*;
        }
        else if(arr*<min2 && arr*!=min1)
            min2=arr*;
        i++;
    }
    //if all nos. are same
    if(min2==1001)
        min2=min1;
    //find gcd of min1 and min2 
    else
    {
       while(min1)
       {
            i=min1;
            min1=min2%min1;
            min2=i;
        }
    }
    i=-1;
    if(min2!=1)
    {
         //check whether gcd divides all nos.
         while(++i<n && !(arr*%min2));
       //  if yes,divide all nos by gcd.
         if(i==n)
         {
             for(i=0;i<n;i++)
                   arr*/=min2;
         }
    }
    //print result
    for(i=0;i<n;i++)
            cout<<arr*<<" ";
    cout<<"

";
}
return 0;
}


#2

@s_enterpreneur

Consider the array : 8 16 22

Minimum(m1)=8 and second minimum(m2)=16.

gcd(m1,m2)=gcd(8,16)=8 which is non-zero.

=>your output is 1 2 2 (note that 22/8 =2)

However required output is 4 8 11.

Hope you understand.