逆序对的数量，主要是利用了归并的思想，借用归并排序就把题解出来了

#include <iostream>

using namespace std;
const int N = 100010;
int q[N], tmp[N];

long long merge_sort(int q[], int l, int r)
{
    if(l >= r) return 0;
    int mid = (l + r)/2;
    long long res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r);
    int i = l, j = mid + 1, k = 0;
    while(i <= mid && j <= r)
    {
        if(q[i] <= q[j]) tmp[k++] = q[i++];
        else 
        {
            tmp[k++] = q[j++];
            res += mid - i + 1;
        }
    }
    while(i <= mid) tmp[k++] = q[i++];
    while(j <= r) tmp[k++] = q[j++];
    for(i = 0, j = l; j <= r; i++, j++)
       q[j] = tmp[i];

    return res;
}
int main()
{
    int i, n;
    scanf("%d", &n);
    for(i = 0; i < n; i++) scanf("%d", &q[i]);
    printf("%ld", merge_sort(q, 0, n - 1));
    return 0;
}