Larry and Inversions

Larry just studied the algorithm to count number of inversions. He's very interested in it. He's considering another problem: Given a permutation of integers from 1 to n, how many inversions it has if we reverse one of its subarray?

Formally speaking, given an integer array a (indices are from 0 to n−1) which contains a permutation of integers from 1 to n, two elements [ and [ form an inversion if [ and i<j. Your job is to count, for each pair of 0, the number of inversions if we reverse the subarray from [ to [.

Input Specification:

Each input file contains one test case. Each case consists of a positive integer n (≤) in the first line, and a permutation of integers from 1 to n in the second line. The numbers in a line are separated by a single space.

Output Specification:

For each test case, output ( integers in a single line. The results are for reversing subarray indicating by all possible pairs of indices 0 in i-major order -- that is, the first n results are for the reverse of subarrary [0..0], [0..1], ...[0..n−1]; the next n−1 results are for the reverse of subarry [1..1], [1..2],..., [1..n−1] and so on.

All the numbers in a line must be separated by a single space, with no extra space at the beginning or the end of the line.

Sample Input:

3
2 1 3

 

Sample Output:

1 0 2 1 2 1

 

Hint:

The original array is { 2, 1, 3 }.

• Reversing subarray [0..0] makes { 2, 1, 3 } which has 1 inversion.
• Reversing subarray [0..1] makes { 1, 2, 3 } which has 0 inversion.
• Reversing subarray [0..2] makes { 3, 1, 2 } which has 2 inversions.
• Reversing subarray [1..1] makes { 2, 1, 3 } which has 1 inversion.
• Reversing subarray [1..2] makes { 2, 3, 1 } which has 2 inversions.
• Reversing subarrays [2..2] makes { 2, 1, 3 } which has 1 inversion.

  #include<bits/stdc++.h>
  using namespace std;
  int n,k;
  vector<int> v;
  vector<int> vl;
  int main()
  {
  //  freopen("data.txt","r",stdin);
        scanf("%d",&n);
        vl.resize(n*n,0);
        for(int i=0;i<n;i++)
        {
          scanf("%d",&k);
          v.emplace_back(k-1);
      }
      int brev=0;
      for(int i=0;i<n;++i)
      {
          int rev=0;
          for(int j=i;j<n;++j)
          {
              if(v[j]<v[i])
              rev++;
              vl[i*n+j]=rev;
          }
          brev+=rev;
      }
      bool flag=false;
      for(int i=0;i<n;i++)
      for(int j=i;j<n;j++)
      {
          if(i==j)
          {
              if(flag)
              printf(" ");
              else
              flag=true;
              printf("%d",brev);
          }
          else
          {
              int drev=0;
              for(int t=i;t<j;t++)
              drev+=vl[t*n+j];
              printf(" %d",brev-2*drev+(j-i)*(j-i+1)/2);
          }
      }
      return 0;
  }