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 <string.h>
#include <limits.h>

#define DEF_ELEM_MAX 200000

typedef long long llong;

int S[DEF_ELEM_MAX];
int L[DEF_ELEM_MAX/2+2];
int R[DEF_ELEM_MAX/2+2];
	int N;
/////////////////////////////////////////////////////////////////////////////////
//
/////////////////////////////////////////////////////////////////////////////////
llong merge(int *A, int left, int mid, int right)
{
	llong cnt = 0;
	int i;
	int j;
	int k;
	int n1 = mid - left;
	int n2 = right - mid;

//  L[0...n1], R[0...n2] ?????????

	for ( i = 0 ; i < n1 ; i++ ) {
		L[i] = A[left + i];
	}

	for ( i = 0 ; i < n2 ; i++ ) {
		R[i] = A[mid + i];
	}

	L[n1] = INT_MAX;
	R[n2] = INT_MAX;
	i = 0;
	j = 0;
	for ( k = left ; k < right ; k++ ) {
		if ( L[i] <= R[j] ) {
			A[k] = L[i++];
		} else {
			A[k] = R[j++];
			cnt += n1 - i; // = mid + j - k -1
		}
	}
	return cnt;
}

/////////////////////////////////////////////////////////////////////////////////
//
/////////////////////////////////////////////////////////////////////////////////
llong mergeSort(int *A, int left, int right)
{
	llong v1;
	llong v2;
	llong v3;
	int mid;
	if (left+1 < right) {
		mid = (left + right)/2;
		v1 = mergeSort(A, left, mid);
		v2 = mergeSort(A, mid, right);
		v3 = merge(A, left, mid, right);
		return v1+v2+v3;
	} else {
		return 0;
	}
}
/////////////////////////////////////////////////////////////////////////////////
//
/////////////////////////////////////////////////////////////////////////////////
int main(void)
{
	llong ans;
	int i;
//
	scanf("%d", &N);
	for ( i = 0 ; i < N ; i++ ) {
		scanf("%d", &S[i] );
	}

	ans = mergeSort(S, 0, N);

	printf("%ld\n", ans);

	return 0;
}