Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

The Number of Inversions

For a given sequence $A = \{a_0, a_1, ... a_{n-1}\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program:

bubbleSort(A)
cnt = 0 // the number of inversions
for i = 0 to A.length-1
for j = A.length-1 downto i+1
if A[j] < A[j-1]
swap(A[j], A[j-1])
cnt++

return cnt

For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded.

Input

In the first line, an integer $n$, the number of elements in $A$, is given. In the second line, the elements $a_i$ ($i = 0, 1, .. n-1$) are given separated by space characters.

output

Print the number of inversions in a line.

Constraints

$ 1 \leq n \leq 200,000$

$ 0 \leq a_i \leq 10^9$

$a_i$ are all different

Sample Input 1

5
3 5 2 1 4

Sample Output 1

6

Sample Input 2

3
3 1 2

Sample Output 2

2

Solution in C# (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

using System.IO;

namespace ALDS1_5_D
{
    class Program
    {
        private static int n;
        private static int[] A;
        private static long pairOfReverse = 0;

        static void Main(string[] args)
        {
            n = int.Parse(Console.ReadLine());
            A = Console.ReadLine()
                       .Split(' ')
                       .Select(s => int.Parse(s))
                       .ToArray();
            Program.MergeSort(0, Program.A.Length - 1);
            Console.WriteLine(Program.pairOfReverse);
        }

        private static void Merge(int leftInd, int midInd, int rightInd)
        {
            int[] leftArry = new int[midInd - leftInd + 1];
            int[] rightArry = new int[rightInd - midInd];

            for (int i = 0; i < leftArry.Length; i++)
            {
                leftArry[i] = Program.A[leftInd + i];
            }

            for (int i = 0; i < rightArry.Length; i++)
            {
                rightArry[i] = Program.A[midInd + 1 + i];
            }

            for (int l = 0, r = 0, k = leftInd; k <= rightInd; k++)
            {
                if (l < leftArry.Length && r >= rightArry.Length || l < leftArry.Length && r < rightArry.Length && leftArry[l] <= rightArry[r])
                {
                    Program.A[k] = leftArry[l];
                    l++;
                }
                else
                {
                    Program.A[k] = rightArry[r];
                    r++;
                    Program.pairOfReverse += leftArry.Length - l;
                }
            }
        }

        private static void MergeSort(int leftInd, int rightInd)
        {
            if (leftInd < rightInd)
            {
                int midInd = (leftInd + rightInd) / 2;
                Program.MergeSort(leftInd, midInd);
                Program.MergeSort(midInd + 1, rightInd);
                Program.Merge(leftInd, midInd, rightInd);
            }
            return;
        }
    }
}