Question

(n+1)(n+2)...2n...n(n-1)(n-2).21逆序数

Answer 1

(n+1)(n+2)...(2n)n(n-1)(n-2).21

我分成两块(n+1)(n+2)...(2n)和n(n-1)(n-2).21

(n+1)(n+2)...(2n)的逆序数和是n*n

n(n-1)(n-2).21的逆序数和是n-1+n-2+n-3+.3+2+1=n(n-1)/2

所以(n+1)(n+2)...(2n)n(n-1)(n-2).21逆序数就是n^2+n(n-1)/2=(3n^2-n)/2