User:Typetive/Template:Check sort C++ code

#include <iostream>
#include <fstream>
#include <time.h>
#include <stdlib.h>
#include <math.h>

inline double Timer() { return ((double)clock())/CLOCKS_PER_SEC; }
inline double Rnd()   { return ((double)rand())/RAND_MAX; }

int main() {
    cout << "Please wait for the RESULTS: (Check sort) is working... " << endl;
    long* a = 0L;
    long* b = 0L;
    double start1, start, finish2, finish1, finish;
    const int one = 1L;
    const int zero = 0L;
    const long maxi = 10000000L;
    const long topnum = maxi;
    const long botnum = one;
    long i = 0L, n = 0L, j = 0L, k = maxi;
    ofstream out("checksort.txt");
    out << endl;
    out.flush();
	delete[] a;
	delete[] b;
	a = new long[k];
	b = new long[k];
		start1 = Timer();
    for(k=botnum; k <= topnum; k += topnum/100) {
        start = Timer();
	for (i=botnum; i <= k; i++) {
	    n = ((int)(k * Rnd())) + botnum;
	    a[n] = one; // (check sort is a(n) = one) 
	}
	finish1 = Timer() - start;
	start = Timer();
	j = 0L;
	for (n = botnum; n <= k; n++) {
	    if (a[n] > zero) { b[j] = n; j = j + one; }
	}
	finish2 = Timer() - start;
	finish = Timer() - start1;
	out << k << "\t" << finish1 << "\t" << finish2 << endl;
	out.flush();
    }
    out.close();
	cout << endl << endl;
	cout << "1.) Total number of items sorted:...................... " << j << endl;
	cout << "2.) Total primary array size to accomodate the sort:... " << maxi << endl; 
	cout << "3.) Total secondary array size to accomodate the sort:. " << j << endl; 
	cout << "4.) Total length of key:............................... " <<
             ( k == 0 ? 0 : (int) ceil( log( (double) k )/log( 10.0 ) ) ) << " decimal digits." << endl; 
	cout << "5.) Total time for setup per 100,000 items:............ " << finish1 << " seconds." << endl; 
	cout << "6.) Total time for sort per 100,000 items:............. " << finish2 << " seconds." << endl; 
	cout << "7.) Total time for program:............................ " << finish << " seconds." << endl; 
        cout << endl << endl;
/*
    for (i=1; i<=j; i++) {
	cout << i << b[i] << endl;
    }
*/
    return 0;
}