How do you say the lower bound is x + 1 ? I understand its x + 1 only when the B = {x} , but the lower bound of the answer should be 2 right ?
Got it Sorry. You can do at least x + 1 .
Yes, got it. Thank you !
I reported this during the competition but apparently it couldn’t be taken down.
why my solution gives run time error instead of running in many test cases successfully
https://www.codechef.com/viewsolution/24964013
Help Needed
I did not understand the part :
Why adding identical elements to B does not affect it but affects C ?
Thanks in advanced 
GCD(8,8,8) = 8 so adding a similar element to a set does not changes the overall GCD.
GCD(6,12,24) = 6 if we add any number of 6 (or any other element currently present in the set) the GCD won’t change.
GCD(6,6,6,6,6,12,24) = 6
so adding identical elements won’t create any difference but if you add that element to C it might create difference.
Thanks again friend 
1 Like
Can anyone help me out pls. Even my solution is failing for first two subcases of subtask -2
https://www.codechef.com/viewsolution/25022048a
I also tried to code this way but it didn’t work. I am not able to understand how any number below the second largest contribute so much to single handedly raise the sum of gcd(even more than the highest element !). Also I am not able to understand what is the need of removing repetitive numbers here. C = {4,4,4,4,4,4,5,5,5,5,5,6,6,6,6} and B = {16} would give same result as C = {4,5,6} and B = {16}
#include <bits/stdc++.h>
#define ll long long int
#define ibs ios_base::sync_with_stdio(false)
#define cti cin.tie(0)
using namespace std;//coded by abhijay mitra
ll solve(std::vector<ll> v,ll n){
sort(v.begin(), v.end());
if (v[0]==v[n-1]) return 2*v[0];
ll h=v[n-1],sh=v[n-1],i=n-1,g1=v[0],g2,g3;
do{
i--;
sh=v[i];
}while(sh==h);
for (int i = 0; v[i] <sh ; ++i)
g1=__gcd(g1,v[i]);
g2=__gcd(g1,h);g3=__gcd(g1,sh);
if(g3+h>g2+sh)
return g3+h;
else return g2+sh;
}
int main()
{
ibs;cti;
int t;cin>>t;cout<<"\n";
while(t--){
ll n;cin>>n;std::vector<ll> v(n);
for (int i = 0; i < n; ++i)
cin>>v[i];
cout<<solve(v,n)<<"\n";
}
return 0;
}
This works
#include <bits/stdc++.h>
#define ll long long int
#define ibs ios_base::sync_with_stdio(false)
#define cti cin.tie(0)
using namespace std;//coded by abhijay mitra
ll solve(std::vector<ll> v,ll n){
sort(v.begin(), v.end());
if (v[0]==v[n-1]) return 2*v[0];
ll h=v[n-1],sh=v[n-1],i=n-1,g1=v[0],g2,g3;
do{
i--;
sh=v[i];
}while(sh==h);
for (int i = 0; v[i] <sh ; ++i)
g1=__gcd(g1,v[i]);
g2=__gcd(g1,h);g3=__gcd(g1,sh);
if(g3+h>g2+sh)
return g3+h;
else return g2+sh;
}
int main()
{
ibs;cti;
int t;cin>>t;cout<<"\n";
while(t--){
ll n;cin>>n;std::vector<ll> v(n);
for (int i = 0; i < n; ++i)
cin>>v[i];
cout<<solve(v,n)<<"\n";
}
return 0;
}
c++ implementation. great idea though
You can read the explanation on stackoverflow. It’ll always work. It took me some times to understand why this works. I just read it and that guy explained it well
https://www.codechef.com/viewsolution/32171611
Pleaase help wa on 2 test cases
please help @waqar_ahmad224
hard to understand the code when it is un-commented.
hard to understand what m1 , m2 and m3 signify.
in the third case why is it max of A+gcd(,B,C…) or B+gcd(A,C…)
we can also take C+gcd(A,B,…) like wise D+gcd(A,B,C,E…) …
why’d you check tho?