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)

#include <stdio.h>

/* Macro Definition */
#define MAX 200000

/* Prototype Declaration */
void merge(int *, int, int, int);
void mergeSort(int *,int, int);

/* Grobal Variable */
int S[MAX], L[(MAX/2) + 1], R[(MAX/2) + 1], n;
long long count = 0;

int main() {
  int i;

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

  mergeSort(S, 0, n);

  printf("%lld\n", count);

  return 0;
}

void merge(int *S, int left, int mid, int right) {
  int i, j = 0, k, n1, n2;
  n1 = mid - left;
  n2 = right - mid;

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

  L[n1] = 1000000500;
  R[n2] = 1000000500;

  i = 0;

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


}

void mergeSort(int *S, int left, int right) {
  int mid;
  
  if( left + 1 < right ) {
    mid = (left + right) / 2;
    mergeSort(S, left, mid);
    mergeSort(S, mid, right);
    merge(S, left, mid, right);
  }

}