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>

#define N 262144
#define HALFN 131072 // 2^17
#define INFTY 1000000001 // 10^9 + 1

int L[HALFN] = {0}, R[HALFN] = {0};
long cnt = 0;

/* void swap(int *x, int *y) { */
/*   int tmp; */

/*   tmp = *x; */
/*   *x = *y; */
/*   *y = tmp; */
/* } */

/* int bubbleSort(int n, int *A) { */
/*   int i, j; */

/*   for (i = 0; i < n; i++) { */
/*     for (j = n - 1; i < j; j--) { */
/*       if (A[j] < A[j-1]) { */
/* 	swap(&A[j], &A[j-1]); */
/* 	cnt++; */
/*       } */
/*     } */
/*   } */
/*   return cnt; */
/* } */
	
void merge(int *A, int left, int mid, int right) {
  int i, j, k, n1, n2;
  
  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] = INFTY;
  R[n2] = INFTY;

  /* // count the number of inversions */
  /* for (j = 0; j < n2; j++) { */
  /*   for (i = 0; i < n1; i++) { */
  /*     if (R[j] < L[i]) { */
  /* 	cnt += (n1 - i); */
  /* 	break; */
  /*     } */
  /*   } */
  /* } */
  
  // sort array
  for (i = 0, j = 0, k = left; k < right; k++) {
    if (R[j] < L[i]) {
      cnt += (n1 - i);
      A[k] = R[j];
      j++;
    }
    else {
      A[k] = L[i];
      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] = {0};

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

  mergeSort(A, 0, n);
  printf("%ld\n", cnt);
  return 0;
}