The goal is to make both numbers, even so that their gcd>1, here 17+1=18 & 29+1=30, total 2 operations.
Oh, got it.
I interpreted the question wrong.
1 Like
why is this logic wrong ?
ll gcd(ll a,ll b)
{
if(a==0)
return b;
return gcd(b%a,a);
}
int main()
{
fio
wit{
ll x,y;
cin>>x>>y;
if(gcd(x,y)>1)
cout<<“0”<<endl;
else if(x%2==0||(y%2==0))
{
cout<<“1”<<endl;
}
else
{ cout<<“2”<<endl;}
}
return 0;
}
Half of the code is missing, so I’ll use CodeChef: Practical coding for everyone, in which case: consider the test input:
1
3 9
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OMG this was a bad mistake , thanks for the help.
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Why 7 27 output is 2 instead of 1?
gcd(7, 28) is 7 > 1
It should be 1 - where are you seeing that the output should be 2?
sorry my bad 
1 Like
though used the same approach but could not get ac. can any one explain what was my mistake?
my code - Fj10CB - Online C++0x Compiler & Debugging Tool - Ideone.com
It fails on the test input:
1
5 21
Try this
1
3 5
Output should be 1 but your code outputs 2
but my code is giving ans 2 for 5 and 21. isn’t correct?
No: 5+1=6, and \gcd(6,21)=3.
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Please Help. Why is this giving WA. Cant think of any case it will fail.
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int main() {
int t;cin>>t;
while(t–)
{
ll x,y,c=0;cin>>x>>y;
while(__gcd(x,y)==1)
{
if(x<y)swap(x,y);
if(x%2==0)y++;
else x++;
c++;
}
cout<<c<<endl;
}
}
Maybe stop running sieve()?
1 Like
17 29 becomes 18 and 30 after 2 operations then gcd(18,30)>1… We can change both values but in different operation.
Didnt’t worked for me…still giving me TLE…pls help
For god sake, watch your code - the indentation sucks. Provide a cleaner code. It becomes easy to find out where it went wrong.