Problem:
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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)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define N 500000
#define INF 1000000000

int L[N], R[N];
long long count = 0;

void merge(int A[], int left, int mid, int right)
{
  int n1, n2, i, j, k;

  n1 = mid - left;
  n2 = right - mid;

  for(i = 0; i < n1; i++) L[i] = A[left + i];
  for(i = 0; i < n2; i++) R[i] = A[mid + i];

  L[n1] = INF;
  R[n2] = INF;

  i = 0;
  j = 0;

  for(k = left; k < right; k++){
    if(L[i] <= R[j]){
      A[k] = L[i];
      i++;
    }
    else{
      A[k] = R[j];
      j++;
      count += n1 - i;
    }
  }
}

void mergeSort(int A[], int left, int right)
{
  int mid;

  if(left + 1 < right){
    mid = (left + right) / 2;
    mergeSort(A, left, mid);
    mergeSort(A, mid, right);
    merge(A, left, mid, right);
  }
}

int main(void)
{
  int i, n, A[N];

  scanf("%d", &n);

  for(i = 0; i < n; i++){
    scanf("%d", &A[i]);
  }

  mergeSort(A, 0, n);

  printf("%lld\n", count);            //%lldとすることでlong long int型の数値をprint文で表示

  return 0;
}