I have a brute-force code for cheking answers
Python 3.8
def summa(a):
    ans = 0
    for i in range(len(a)):
        ans += a[i] * a[i - 1]
    return ans
def recursion(n, do_etogo):
    a = True
    mn = 10 ** 9
    mx = 0
    mnansn = []
    mxansn = []
    for i in range(2, n + 1):
        if i in do_etogo:
            continue
        a = False
        mnans, mxans, mnansm, mxansm = recursion(n, do_etogo + [i])
        if mnans < mn:
            mn = mnans
            mnansn = [mnansm]
        elif mnans == mn:
            mnansn.append(mnansm)
        if mxans > mx:
            mx = mxans
            mxansn = [mxansm]
        elif mxans == mx:
            mxansn.append(mxansm)
    if a:
        return summa(do_etogo), summa(do_etogo), do_etogo, do_etogo
    return mn, mx, mnansn, mxansn
n = int(input())
print(recursion(n, [1]))

Платит, куда денется =)

Write brute-force algo and you'll find how does optimal solution look like...

Check it yourself)) Smth like if (n > 50) while (1);

write brute force program and you will find your mistake

I've understood what's the metter.
input: 5
possible outputs:
1 5 2 4 3
1 3 5 4 2
or
1 4 3 2 5
1 2 4 5 3
But only the second solution is accepted.

Edited by author 25.03.2007 21:09

But better of course - min & max )

You are wrong. Read the task more carefully.

You are wrang, becouse 1*4+4*2+2*3+3*1 = 21, 4*1+1*2+2*3+3*4 = 24.

1*4+4*2+2*3+3*1=21
1*3+3*2+2*4+4*1=21

For minimal permutation must be:
(p[i] - p[i+1]) * (p[i-1] - p[i+2]) < 0