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 <math.h>
#include <string.h>
#define rad M_PI/3
#define N 500001

int S[N]={0};
int L[N/2]={0};
int R[N/2]={0};
long count =0;

void Merge(int A[],int left,int mid,int right){
    int n1,n2;
    int i,j,k;
    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]=1000000010;
    R[n2]=1000000010;
    i=j=0;
    for(k=left;k<right;++k){
        if(L[i]<=R[j]){
            A[k]=L[i];
            ++i;
    }else{
        A[k]=R[j];
        ++j;
        count+=(n1-i);
    }
    }
    return;
}
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);
    }
    return;
}

int main()
{
  int n, i;
 
    scanf("%d", &n);
 
    for (i = 0; i < n; ++i) {
        scanf("%d", &S[i]);
    }
    MergeSort(S, 0, n);
    printf("%ld\n", count);
     
    return 0;
  return 0;
}