// program illustrating use of vectors.    bds 2004Mar

/* Input is a sequence of numbers.
   Output is the same sequence of numbers, in the same order, and with
   each number on a new line.  Also the largest number is to be indicated
   by the word "largest" appearing after the number.  Similarly for the 
   smallest.
*/
#include <iostream>
#include <vector>
using namespace std;

// comments beginning with //// mark first use of each vector functionality

int main()
{
    vector<float> A;       //// default constructor - empty vector
    float x;

// input the numbers
    cin >> x;
    while (cin) 
    {   
	A.push_back(x);    //// push_back()  - grow size by one, put x in.
	cin >> x;
    }

    if ( A.size() == 0 ) return 0;         
			   //// size() - returns an int, current size.

// find the position of the max  (illustrating index use)

    float maxval = A[0];
			   //// operator[]  (indexing)
			   //// relation to int as index type
    int i, maxpos = 0; 
			   //// int default constructor 
			   //// int copy constructor

    for( i = 1; i < A.size(); ++i )
	if ( A[i] > maxval ) { maxval = A[i]; maxpos = i; }
	
// find the position of the min (illustrating iterator use)

    float minval = A.front();
			   //// front() - first element, a float
    vector<float>::iterator j, minptr = A.begin(); 
			   //// ::iterator - a type "owned" by the vector type
			   //// ::iterator's default constructor
			   //// begin() - returns an ::iterator pointing to front
			   //// ::iterator copy constructor 


    for ( j = A.begin() + 1; j < A.end(); ++j )
			   //// end() - returns an ::iterator poiting beyond end
			   //// ::iterator operator+  - add an int to it.
			   //// ::iterator assignment 
			   //// ::iterator operator<  - compare positions
			   //// ::iterator operator++ - point to next posn.
	if ( *j < minval ) { minval = *j; minptr = j; }
			   //// ::iterator operator* - returns the value being
		 	   //// 		       pointed to, a float
			   //// ::iterator operator= - ::iterator assignment
	

// output the result

    for ( j = A.begin(); j < A.end(); ++j )
    {
	cout << *j;
	if ( minptr == j ) cout << " smallest";
	if ( maxpos == j - A.begin() ) cout << " largest";
			   //// ::iterator operator- - difference is an int.
	cout << endl;
    }
	
    return 0;
}