*[codility]ArrayInversionCount

http://codility.com/demo/take-sample-test/arrayinversioncount

求逆序对数，归并排序并记录逆序次数。

// you can also use includes, for example:
// #include <algorithm>
int merge(vector<int> &A, int left, int right) {
    if (left >= right) return 0;
    int mid = left + (right - left) / 2;
    int left_count = merge(A, left, mid);
    int right_count = merge(A, mid + 1, right);
    vector<int> tmp;
    int i = left;
    int j = mid + 1;
    int merge_count = 0;
    while (i <= mid || j <= right) {
        if (i <= mid && j <= right) {
            if (A[i] > A[j]) {
                tmp.push_back(A[i++]);
                merge_count += (right - j + 1);
            }
            else {
                tmp.push_back(A[j++]);
            }
        }
        else if (i <= mid) {
            tmp.push_back(A[i++]);
        }
        else {
            tmp.push_back(A[j++]);
        }
    }
    for (int k = 0; k < tmp.size(); k++) {
        A[left + k] = tmp[k];
    }
    return (left_count + right_count + merge_count);
}

int solution(const vector<int> &A) {
    // write your code in C++98
    vector<int> B(A);
    return merge(B, 0, B.size() - 1);
}