Assume you have a int variable n that has already been declared and initialized. Its value is the number of integers that need to be read in from standard input and printed out in sorted (ascending) order, each on a line by itself. Furthermore, there are no duplicates in the input and every number to be read is a non-negative value that is less than n’s value. In this exercise you may not use any array (or fancy STL collection such as a vector). You may declare a variable or two as needed. With these restrictions, read the n values and print them out as required onto standard output.

Assume you have a int variable n that has already been declared and initialized. Its value is the number of integers that need to be read in from standard input and printed out in sorted (ascending) order, each on a line by itself. Furthermore, there are no duplicates in the input and every number to be read is a non-negative value that is less than n’s value.  In this exercise you may not use any array (or fancy STL collection such as a vector). You may declare a variable or two as needed. With these restrictions, read the n values and print them out as required onto standard output.

LANGUAGE: C++

CHALLENGE:

Assume you have a int variable n that has already been declared and initialized. Its value is the number of integers that need to be read in from standard input and printed out in sorted (ascending) order, each on a line by itself. Furthermore, there are no duplicates in the input and every number to be read is a non-negative value that is less than n’s value.

In this exercise you may not use any array (or fancy STL collection such as a vector). You may declare a variable or two as needed. With these restrictions, read the n values and print them out as required onto standard output.

SOLUTION:


int x=0;

for (int i=0; i<n; i++){
    cin >> x;
    cout << i << "\n";
}

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