Dev Skill DCP-135: Abundant Number Solution in C++
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Dev Skill DCP-135: Abundant Number Solution in C++
Dev Skill DCP-135: Abundant Number Solution in C++
#include<bits/stdc++.h>
using namespace std;
long divisor(long a)
{
long s=1,i,j;
for(i=2;i<a;i++)
{
if(a%i==0)
{
s=s+i;
}
}
if(a<s)
return s-a;
else
return 0;
}
int main()
{
int t,n;
cin>>t;
while(t--)
{
long a,ab=0,ba=0,c;
cin>>n;
while(n--)
{
cin>>a;
ab=divisor(a);
if(ab>ba)
{
ba=ab;
c=a;
}
}
if(ba>0)
cout<<c<<" "<<ba<<endl;
else
cout<<"-1"<<endl;
}
return 0;
}
using namespace std;
long divisor(long a)
{
long s=1,i,j;
for(i=2;i<a;i++)
{
if(a%i==0)
{
s=s+i;
}
}
if(a<s)
return s-a;
else
return 0;
}
int main()
{
int t,n;
cin>>t;
while(t--)
{
long a,ab=0,ba=0,c;
cin>>n;
while(n--)
{
cin>>a;
ab=divisor(a);
if(ab>ba)
{
ba=ab;
c=a;
}
}
if(ba>0)
cout<<c<<" "<<ba<<endl;
else
cout<<"-1"<<endl;
}
return 0;
}
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