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>

#define MAX 200000
#define SENTINEL 1000000001

int gA[MAX+1],gn,gcnt;
int gL[ (MAX/2)+1 ],gR[ (MAX/2)+1 ];
 
void merge( int left, int mid, int right )
{
    int i,j,k;
    int n1=mid-left;
    int n2=right-mid;
 
    for( i=0; i<n1; i++ ) {
        gL[i]=gA[ left+i ];
    }
    for( i=0; i<n2; i++ ) {
        gR[i]=gA[ mid+i ];
    }
     
    gL[n1]=gR[n2]=SENTINEL;
    i=j=0;
    for( k=left; k<right; k++ ) {
        if( gL[i] <= gR[j] ) {
            gA[k]=gL[i++];
        } else {
	        gcnt+=(n1-i);
            gA[k]=gR[j++];
        }
    }
    return;
}
 
void mergeSort( int left, int right )
{   
    int mid;
    if( left+1 < right ) {
        mid=(left+right)/2;
        mergeSort( left, mid );
        mergeSort( mid, right );
        merge( left, mid, right );
    }
     
    return;
}


int main( void ) 
{	
	int i;
	
    scanf( "%d", &gn );

	for( i=0; i<gn; i++ ) {
		scanf( "%d", &gA[i] );
	}
	gcnt=0;
	mergeSort( 0, gn );
	printf( "%d\n",  gcnt );

    return 0;
}