AtCoder Beginner Contest 190
F
Shift and Inversions
AtCoder
⌀ 1321
原题
← E→
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Problem Statement

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Given is a sequence $A = [a_0, a_1, a_2, \dots, a_{N-1}]$ that is a permutation of $0, 1, 2, \dots, N - 1$. For each $k = 0, 1, 2, \dots, N - 1$, find the inversion number of the sequence $B = [b_0, b_1, b_2, \dots, b_{N-1}]$ defined as $b_i = a_{i+k \bmod N}$.

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What is inversion number? The inversion number of a sequence $A = [a_0, a_1, a_2, \dots, a_{N-1}]$ is the number of pairs of indices $(i, j)$ such that $i < j$ and $a_i > a_j$.

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Constraints

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  • All values in input are integers.
  • $2 ≤ N ≤ 3 \times 10^5$
  • $a_0, a_1, a_2, \dots, a_{N-1}$ is a permutation of $0, 1, 2, \dots, N - 1$.
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Input

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Input is given from Standard Input in the following format:

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N
a_0 a_1 a_2 \cdots a_{N-1}

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Output

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Print $N$ lines. The $(i + 1)$-th line should contain the answer for the case $k = i$.

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样例

Input #1
4
0 1 2 3
Output #1
0
3
4
3
We have $A = [0, 1, 2, 3]$. For $k = 0$, the inversion number of $B = [0, 1, 2, 3]$ is $0$. For $k = 1$, the inversion number of $B = [1, 2, 3, 0]$ is $3$. For $k = 2$, the inversion number of $B = [2, 3, 0, 1]$ is $4$. For $k = 3$, the inversion number of $B = [3, 0, 1, 2]$ is $3$.
Input #2
10
0 3 1 5 4 2 9 6 8 7
Output #2
9
18
21
28
27
28
33
24
21
14
题目信息
题号
abc190_f
时间限制
2 s
内存限制
1024 MB

题面与样例版权归 AtCoder 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。