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## Discussion of Problem 1118. Nontrivial Numbers

AC in 0.015
Posted by Aditya Singh 15 Apr 2017 05:24
#include<bits/stdc++.h>
using namespace std;
bool B[1000008];
float check(long N)
{
long K=sqrt(N);
long D=1;
for(long i=2;i<=K;i++)
if(N%i==0)
D+=(i+N/i);
return (D/(float)(N));
}
void prime()
{
for(long i=2;i<=1000;i++)
if(B[i]==0)
for(long j=2;i*j<=1000000;j++)
B[i*j]=1;
}
int main()
{
ifstream cin("input.in");
ofstream cout("output.out");
long X,Y;
cin>>X>>Y;
prime();
if(X==1)
cout<<1;
else{
bool flag=0;
for(long i=Y;i>=X;i--)
if(B[i]==0){
cout<<i;
flag=1;
break;
}
if(!flag){
float P=2e9;
long pos;
for(long i=X;i<=Y;i++)
{
float H=check(i);
if(H<P){
pos=i;
P=H;
}
}
cout<<pos;
}
}
}
Re: AC in 0.015
Posted by Igor Kulatov 10 Feb 2018 15:29
Hi,

Very good solution! Just one note, it doesn't affect AC, but:

In D+=(i+N/i); if (i == N/i) you don't need to add both i and N/i, but just i.

In this case result of check(N) will be 100% equal to triviality(N) in all cases.