@atul:
Anurag's code has illustrated the idea of O(sqrt(E)) solution.
One more thing to optimize is that we can safely break after finding the
first factor of E which is less than sqrt(E).
So the code could be changed to:
#include<cmath>#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>using
namespace std;#define INFY 1000000007int main()
{
int n, i, j;
int val, minDiv, minDis;
while(1)
{
cin >> n;
minDis = INFY;
for (i = n; i <= n+2; i++)
{
*for (j = sqrt(i * 1.0); j > 0; j--)*
{
if (i % j == 0 && minDis > (i/j - j) )
{
minDis = i/j - j;
minDiv = j;
val = i;
*break;*
}
}
}
cout<<val<<" "<<minDiv<<" "<<val/minDiv<<endl;
}
//system("pause");
return 0;
}
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