c++ - C++ マージソートの実装

Question

C++とベクトルでマージソートを実装したいです（整数の入力数が不明であることを考慮して）。ウィキペディアで考えたボトムアップ実装に基づいて実装しようとしました: http://en.wikipedia.org/wiki/Merge_sort。次のコードが出てきました。コンパイルは通過できますが、数値を入力すると終了します。ここで何が問題なのかわかりません。

#include<iostream>
#include<vector>
using namespace std;

void BottomUpMerge(vector<int>, vector<int>::iterator, vector<int>::iterator, vector<int>::iterator, vector<int>);

void BottomUpSort(int n, vector<int> A, vector<int> B)
{
  int width;

  /* Each 1-element run in A is already "sorted". */

  /* Make successively longer sorted runs of length 2, 4, 8, 16... until whole array is sorted. */
  for (width = 1; width < n; width = 2 * width){
    vector<int>::iterator leftstart;

    /* Array A is full of runs of length width. */
    for (leftstart = A.begin(); leftstart <= A.end(); leftstart = leftstart + 2 * width){
        /* Merge two runs: A[i:i+width-1] and A[i+width:i+2*width-1] to B[] */
        /* or copy A[i:n-1] to B[] ( if(i+width >= n) ) */
        vector<int>::iterator rightstart;
        vector<int>::iterator rightend;
        if(leftstart+width >= A.end()){
            rightstart = A.end();
        }
        else{
            rightstart = leftstart+width;
        }
        if(leftstart+2*width >= A.end()){
            rightend = A.end();
        }
        else{
            rightend = leftstart+2*width;
        }

        BottomUpMerge(A, leftstart, rightstart, rightend, B);
    }

    /* Now work array B is full of runs of length 2*width. */
    /* Copy array B to array A for next iteration. */
    /* A more efficient implementation would swap the roles of A and B */
    A = B;
    /* Now array A is full of runs of length 2*width. */
    }
}

void BottomUpMerge(vector<int> A, vector<int>::iterator iLeft, vector<int>::iterator iRight, vector<int>::iterator iEnd, vector<int> B)
{
  vector<int>::iterator i0 = iLeft;
  vector<int>::iterator i1 = iRight;
  vector<int>::iterator j = B.begin() + distance(iLeft, A.begin()); //iterator for vector B

  /* While there are elements in the left or right lists */
  for (; j != B.end(); j++){
    /* If left list head exists and is <= existing right list head */
    if (i0 < iRight && (i1 >= iEnd || *i0 <= *i1)){
        *j = *i0;
        i0 = i0 + 1;
    }
    else{
        *j = *i1;
        i1 = i1 + 1;
    }
  }
}


int main(){
    int input;
    vector<int> numbers;    //input vector of numbers

    cout << "Please input the numbers to be sorted:" << endl;
    while(cin>>input){
        numbers.push_back(input);
    }

    int length = numbers.size();
    vector<int> sorted = numbers;   //work vector ensure has the same size of the original vector

    BottomUpSort(length, numbers, sorted);

    vector<int>::iterator it;
    for(it = numbers.begin(); it != numbers.end(); ++it) {
        std::cout << (*it) << '\n';
    }

    system("pause");
    return 1;
}