//  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>
#include <cmath>
using namespace std;

#include "timer.h"


void
task( ) 
{
    // any constant time task.
    // this one calculates some weird trig stuff
	double x = 0.0;
	double dx = 0.1234;
	double sum = 0.0;
	while( x < 1.0 ) 
    {
		sum += sin( x ) + cos( x );
		x += dx;
	}

    return;
}


int
main( ) 
{
	double const minTime = 2.0;

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

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

		//  This is the code segment being timed.

			for( long i=1; i<=n; ++i ) 
            {
				for( long j=1; j<=n; ++j ) 
                {
					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;
}