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: Wrong Answer)

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

/*
int bubbleSort(int A[], int n) {
	int i, j, t, s;
	int cnt = 0; // 反転数
	for(i=0;i<n-1;i++) {
		for(j=n-1,s=i+1;j>=s;j--) {
	    	if(A[j] < A[j-1]) {
	    		t = A[j];
	    		A[j] = A[j-1];
	    		A[j-1] = A[j];
				cnt++;
	    	}
		}
	}
  
	return cnt;
}
*/

long long int A[200005];
int merge_count(int l, int r){
	int count = 0;
	int i, j, k;
	int n = r - l;
	if(n > 1){
		long long int b[n+5];
		int m = l + n/2;
		count += merge_count(l, m);
		count += merge_count(m, r);
		memcpy((void*)b, (void*)&A[l], sizeof(b));
		for(i = l, j = 0, k = n/2; i < r; ++i){
			if(j == n/2) {
				A[i] = b[k++];
			} else if(k == n) {
				A[i] = b[j++];
			} else if(b[j] <= b[k]) {
				A[i] = b[j++];
			} else {
				A[i] = b[k++];
				count += n/2 - j;
			}
		}
	}
	return count;
}

int bubble_sort(int n) {
	return merge_count(0, n);
}

int main(void) {
	int i, n;
	scanf("%d\n", &n);
	
	for(i=0;i<n;i++) {
		scanf("%lld", &A[i]);
	}
	
	printf("%d\n", bubble_sort(n));

	return 0;
}