//  Time a simple nested loop program segment.
//  Input is the size of the problem instance, n

//  In order to time very short segments the segment is repeated until
//  the total time is at least minTime.  The total time is divided by
//  the number of repetitions to estimate the time for one segment run.

//  written  16 jun 98  tom bailey

#include <iostream>
using namespace std;

#include "timer.h"


void
task( ) 
{
    // any constant time task.
    // this one just returns.
    return;
}


int
main( ) 
{
	double const minTime = 2.0;

	long n;
	cout << "Enter n ( <= 0 to halt ): ";
	cin >> n;
	while( n>0 ) 
    {
		Timer watch;

		int runs = 0;
		while( watch.time() < minTime ) 
        {
			watch.start();

		//  This is the code segment being timed.

			for( int i=1; i<=n; ++i ) 
            {
				task();
			}

  		//  End of timed code segment.

			watch.stop();
			runs++;
        }

		double time = watch.time() / runs;

		cout << endl << "n = " << n 
			<< ",  time = ( " << runs << " runs @ "
			<< time << " seconds per run )" << endl;

		cout << endl << "Enter n ( <= 0 to halt ): ";
		cin >> n;
	}

	return 0;
}