归并排序-分治法求序列逆序对数量

核心思想
1. 序列的逆序对可以分成3类，两个数都在mid 左边，两个数都在mid右边，或者一左一右
2. 对于一左一右的，根据归并排序的过程，可以很容易求的
3. 最后返回总的逆序对数量即可

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010;
int q[N], tmp[N];
typedef long long LL;  // ⚠️可能会爆int


LL merge_sort(int *q, int l, int r) {
    if (l >= r) return 0;
    int mid = (l + r) >> 1;
    int k = 0, i = l, j = mid + 1;
    LL res = merge_sort(q, l, mid);
    res += merge_sort(q, mid + 1, r);

    while (i <= mid && j <= r) {
        if (q[i] <= q[j]) tmp[k++] = q[i++];
        else {
            tmp[k++] = q[j++];
            res += mid - i + 1;  // 当q[i] > q[j] 时，那么i～mid中间的这些数均大于q[j]
        }
    }
    while (i <= mid) tmp[k++] = q[i++];
    while (j <= r) tmp[k++] = q[j++];

    for (int i = l, j = 0; i <= r; i++, j++) q[i] = tmp[j];
    return res;
}

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> q[i];
    cout << merge_sort(q, 0, n - 1) << endl;

    return 0;
}