#P9843. [ICPC2021 Nanjing R] Paimon Sorting

[ICPC2021 Nanjing R] Paimon Sorting

题目描述

Paimon just invents a new sorting algorithm which looks much like bubble sort\textit{bubble sort}, with a few differences. It accepts a 11-indexed sequence AA of length nn and sorts it. Its pseudo-code is shown below.

// The Sorting Algorithm
SORT(A)
  for i from 1 to n // n is the number of elements if A
    for j from 1 to n
      if a[i] < a[j] // a[i] is the i-th element in A
        Swap a[i] and a[j]

If you don't believe this piece of algorithm can sort a sequence it will also be your task to prove it. Anyway here comes the question:

Given an integer sequence A=a1,a2,,anA = a_1, a_2, \cdots, a_n of length nn, for each of its prefix AkA_k of length kk (that is, for each 1kn1 \le k \le n, consider the subsequence Ak=a1,a2,,akA_k = a_1, a_2, \cdots, a_k), count the number of swaps performed if we call SORT(Ak)\text{SORT}(A_k).

输入格式

There are multiple test cases. The first line of the input contains an integer TT indicating the number of test cases. For each test case:

The first line contains an integer nn (1n1051 \le n \le 10^5) indicating the length of the sequence.

The second line contains nn integers a1,a2,,ana_1, a_2, \cdots, a_n (1ain1 \le a_i \le n) indicating the given sequence.

It's guaranteed that the sum of nn of all test cases will not exceed 10610^6.

输出格式

For each test case output one line containing nn integers s1,s2,,sns_1, s_2, \cdots, s_n separated by a space, where sis_i is the number of swaps performed if we call SORT(Ai)\text{SORT}(A_i).

Please, DO NOT output extra spaces at the end of each line or your solution may be considered incorrect!

题目大意

给出一个排序算法(用伪代码表示):

// 排序算法
SORT(A)
  for i from 1 to n // n 是序列 A 的元素个数
    for j from 1 to n
      if a[i] < a[j] // a[i] 是序列 A 的第 i 个元素
        Swap a[i] and a[j]

请你算出对于一个序列 A=a1,a2,,anA=a_1,a_2,\cdots,a_n 的所有前缀 Ak=a1,a2,,akA_k=a_1,a_2,\cdots,a_k1kn1\le k\le n),SORT(Ak)\operatorname{SORT}(A_k) 中的交换(Swap)操作将会被执行几次。

3
5
2 3 2 1 5
3
1 2 3
1
1

0 2 3 5 7
0 2 4
0