本来在想其他的题，好像想偏了，不过最后总结找到了标题所对应的公式。

 

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 10000;
#define lowbit(x) ((x)&(-x))
int n, x;
int c[maxn], a[maxn];
void update(int x, int v)
{
    for(int i = x; i <= n; i += lowbit(i))
        c[i] += v;
}
int  getsum(int x)
{
    int ans = 0;
    for(int i = x; i > 0; i -= lowbit(i))
        ans += c[i];
    return ans;
}
int main()
{
    int ans = 0;
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i = 1; i <= n; i++)
        a[i] = i;
    do{
        memset(c, 0, sizeof(c));
        printf("当前排列：");
        for(int i = 1; i <= n; i++)
            cout << a[i] << " ";
        cout << endl;
        int sum = 0;
        for(int i = 1; i <= n; i++)
        {
            update(i, 1);
            sum += getsum(n) - getsum(a[i]);
        }
        printf("逆序对数目：%d\n", sum);
        ans += sum;
    }while(next_permutation(a+1, a+1+n));
    cout << ans << endl;
    return 0;
}