In this exercise, you will write some code that reads n unique (no duplicates!)

Question

In this exercise, you will write some code that reads n unique (no duplicates!) non-negative integers , each one less than  fifty (50). Your code will print them in sorted order without using any nested loops-- potentially very efficient! We'll walk you through this:

First, assume  you are given  an int variable  n, that contains the number of integers  to read from standard input.

Also assume  you are given  an array , named  wasReadIn, of fifty (50) bool elements  and initialize  all the elements  to false .

Third, read in the n integers  from the input, and each time you read an integer , use it as an index into the bool array , and assign  that element  to be true -- thus "marking" in the array  which numbers have been read.

Lastly the "punchline": write a loop that traverses the bool array : every time it finds an element  that is true  it prints out the element 's INDEX -- which was one of the integers  read in. Place all the numbers on a single line, separated by a single spaces. Note: this technique is not limited to 50 elements -- it works just as well for larger  values . Thus, for example you could have an array  of 1,000,000 elements  (that's right-- one million!) and use it to sort numbers up to 1,000,000 in value !

Explanation / Answer

#include <iostream>
using namespace std;

int main()
{
int x, n=0;
bool wasReadIn[50];
for (int i=0; i<50; i++)
wasReadIn[i] = false;
while(n<1 || n>50)
{
cout << "Numbers to enter: ";
cin >> n;
}
for (int i=0; i<n; i++)
{
cout << "Enter num: ";
cin >> x;
if (!wasReadIn[x] && x >0 && x < 50)
wasReadIn[x] = true;
else{
i--;
cout << "BAD DOG BAD"<<endl;
}
}

for (int i=0; i<50; i++)
{
if (wasReadIn[i])
cout << i << " ";
}
return 0;
}

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