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recursion and printf i guess....

have a static variable and keep increment, print them from constructor . create array of 100 objects which will print 100. :-)

100 printf statements in the simplest case. O(1) algorithm !

#include<iostream>
using namespace std;

void Print( int n )
{
        cout << n;
        if(n<100)
        Print(n+1);
}

int main()
{
        Print(0);
return 0;
}

It's actually o(n) isn't it

one printf to write 1,2,3,4,5....100 or 100 printf will take O(1) but of no use
recursion is the only correct way to print this, I guess.

public static int print(int n){
if (n==0){
return n+1;
}
print(n-1);
System.out.println(n);
return n;
}

public class OutputWithoutLoop
{
public static void main(String[] args)
{
output(100);
}

public static void output(int number)
{
while(number > 0)
{
System.out.println(" number is :" + number);
number--;
}
}
}

void print1ToN(int n){
		if(n > 1){
			print1ToN(n-1);
		}
		System.out.println(n);
	}