解题思路

枚举所有的可能的交换情况，时间复杂度 $O(n^4)$。

用归并排序计算数组的逆序对，时间复杂度 $O(n)$。

综上时间复杂度 $O(n^5)$。

由于 Python 运行效率较低，约 $500$ 秒可得到结果。

N = 55

n = 51
a = [0] * N
tmp = [0] * N

def merge_sort(q, l, r):
    if l >= r:
        return 0
    
    mid = (l + r) // 2
    res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r)
    
    i = l
    j = mid + 1
    k = 0
    while i <= mid and j <= r:
        if q[i] <= q[j]:
            tmp[k] = q[i]
            k += 1
            i += 1
        else:
            tmp[k] = q[j]
            res += mid - i + 1
            k += 1
            j += 1
    while i <= mid:
        tmp[k] = q[i]
        k += 1
        i += 1
    while j <= r:
        tmp[k] = q[j]
        k += 1
        j += 1
    
    for i in range(l, l + k):
        q[i] = tmp[i - l]
    
    return res

cnt = 0
res = 0
for i in range(1, n + 1):
    for j in range(1, n + 1):
        if i != j:
            for u in range(1, n + 1):
                for v in range(1, n + 1):
                    if u != v:
                        cnt += 1
                        a = list(range(0, n + 1))
                        a[i], a[j] = a[j], a[i]
                        a[u], a[v] = a[v], a[u]
                        res += merge_sort(a, 1, n)

print("%.2lf" % (res / cnt))

运行结果：

65.33

【在线测评】