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<limits.h>

int INF = 999999999;

long long int merge(int S[], int left, int mid, int right){
	int i=0, j=0;
	long long int count=0;
	int n1 = mid-left;
	int n2 = right - mid;
	int L[n1], R[n2];
	
	for(i=0; i<n1; i++){
		L[i] = S[left + i];
	}
	for(i=0; i<n2; i++){
		R[i] = S[mid + i];
	}
	L[n1] = INF;
	R[n2] = INF;
	i=0; j=0;
	for(int k=left; k<right; k++){
		if(L[i]<=R[j]){
			S[k] = L[i];
			i++;
		}
		else{
			S[k] = R[j];
			j++;
			count += n1 - i;
		}
	}
	return count;
}

long long int mergeSort(int S[], int left, int right){
	int mid;
	long long int v1, v2, v3;
	mid = (left + right)/2;
	if( left+1<right ){
		mid = (left + right)/2;
		v1 = mergeSort(S, left, mid);
		v2 = mergeSort(S, mid, right);
		v3 = merge(S, left, mid, right);
		return v1 + v2 + v3;
	}
	else return 0;
}

int main(void){
	int n, i, S[500000];
	long long sum;
	scanf("%d", &n);
	for(i=0; i<n; i++){
		scanf("%d", &S[i]);
	}
	sum = mergeSort(S, 0, n);
	
	printf("%lld\n", sum);
	return 0;
}