Suitor elimination - c++

Question

I am trying to teach myself C++ and I came across this program project in my book I am working from:
In an ancient land, the beautiful princess Eve had many suitors. She decided on the following procedure to determine which suitor she would marry. First, all of the suitors would be lined up one after the other and assigned numbers.
The first suitor would be number 1, the second number 2, and so on up to the last suitor,number n. Starting at the first suitor she would then count three suitors down the line (because of the three letters in her name) and the third suitor would be eliminated from winning her hand and removed from the line. Eve would then continue, counting three more suitors, and eliminating every third suitor. When she reached the end of the line she would continue counting from the beginning.
For example, if there were six suitors then the elimination process would proceed as follows:
123456 initial list of suitors, start counting from 1
12456 suitor 3 eliminated, continue counting from 4
1245 suitor 6 eliminated, continue counting from 1
125 suitor 4 eliminated, continue counting from 5
15 suitor 2 eliminated, continue counting from 5
1 suitor 5 eliminated, 1 is the lucky winner
Write a program that uses a vector to determine which position you should stand in to marry the princess if there are n suitors. You will find the following function from the Vector class useful:
v.erase(iter);
// Removes element at position iter
For example, to use this function to erase the fourth element from the beginning of a vector variable named theVector , use
theVector.erase(theVector.begin( ) + 3);
The number 3 is used because the first element in the vector is at index position 0.
I have some preliminary code written, but I am having a hard time figuring out how to tell the program after the first suitor (i.e the 3rd suitor) is eliminated to start counting from the fourth suitor, and so on. Perhaps a nested loop would work? I have found solutions online that use a class but it is difficult for me to understand and I feel like there is a simpler way of solving this problem, any help would be greatly appreciated.
#include <iostream>
#include <vector>
using namespace std;
int main ()
{
int n;
vector<int> vec;
cout << "Enter the number of suitors: " << endl;
cin >> n;
// set some values (from 1 to n)
for(int i = 0; i <= n; i++){
vec.push_back(i);
}
// erase third suitor
vec.erase(vec.begin()+2);
// print vector with erased suitor
for(unsigned i = 0; i <= vec.size(); i++){
cout << vec[i] << endl;
}
}

Answer 1

First of all, there are two bugs in your program: you're initially putting [0, n] in the vector, which should be [1, n], and when printing the contents of the vector, you use <= where it should be <.
Now for the actual question. We want to iterate over the vector with steps of size 2 (the second person to the right of the current person):
for (int i = 0; i < vec.size(); i += 2)
However, when we reach the end of the array, we want to continue counting from the front. For this, we can use the modulo operator %:
for (int i = 0; i < vec.size(); i = (i + 2) % vec.size())
This will restrict i to the range [0, vec.size() - 1]. As such, our loop condition is now useless. Instead, we need to take care to terminate the loop when the vector's size is 1:
for (int i = 0; vec.size() > 1; i = (i + 2) % vec.size())
Putting it together, we get the following:
for (int i = 0; vec.size() > 1; i = (i + 2) % vec.size())
vec.erase(vec.begin() + i + 2);
Or equivalently:
for (int i = 2; vec.size() > 1; i = (i + 2) % vec.size())
vec.erase(vec.begin() + i);
The only element that is in the vector when this loop terminates is the number of the lucky suitor.
EDIT: To print out the contents of the vector after each elimination, try this:
for (int i = 2; vec.size() > 1; i = (i + 2) % vec.size())
{
vec.erase(vec.begin() + i);
for (int j = 0; j < vec.size(); j++)
cout << vec[j];
cout << endl;
}

Answer 2

Use relative positioning and take advantage of vec.erase()'s return value.
newpos = vec.erase(pos);
Here 'newpos' is pointing to the element that followed the erased one. Meaning, earsing '3' from {1, 2, 3, 4, 5, 6} gets you pointed to '4'. You can then do
pos = vec.begin();
while (pos != vec.end()) {
erasePos = // compute erasing position relative to 'pos'
// e.g. pos + 2, within bounds
pos = vec.erase(erasePos);
}
You need to do this in a loop.

Answer 3

For that use case where you simply iterate until the end and loop back to the beginning, I would use a list instead of a vector. The program could be :
#include <iostream>
#include <list>
using namespace std;
int main() {
list<int> l;
int n;
int i;
cout << "Enter the number of suitors: ";
cin >> n;
for (i=0; i<n; i++) l.push_back(i);
int delta = 3; // want to remove every third
i = delta;
list<int>::iterator it = l.begin();
while (l.size() > 1) { // loop until only one left
if (--i == 0) { // is it third ?
list<int>::iterator it2 = it++; // move pointer one step further (erase would destroy it)
l.erase(it2); // remove it
i = delta; // rearm counter
}
else it++; // was not third, simply goes on
if (it == l.end()) it = l.begin(); // if at end, go back to begin
}
cout << "Winner is number : " << l.front() + 1 << endl; // add one as our list was 0,1,...n-1
return 0;
}