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>
#define MAX 200000
#define SENTINEL 2000000000
typedef long long llong;

int L[MAX/2+2],R[MAX/2+2];

llong merge(int*,int,int,int,int);
llong mergeSort(int*,int,int,int);

int main(){
  int A[MAX],n,i;

  scanf("%d",&n);

  for(i=0;i<n;i++){
    scanf("%d",&A[i]);
  }

  llong ans=mergeSort(A,n,0,n);

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

  return 0;
}

llong merge(int *A,int n,int left,int mid,int right){
  int i,j,k;
  llong cnt=0;
  int n1=mid-left;
  int 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]=R[n2]=SENTINEL;
  i=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;
    }
  }

  return cnt;
}

llong mergeSort(int *A,int n,int left,int right){
  int mid;
  llong v1,v2,v3;

  if(left+1<right){
    mid=(left+right)/2;
    v1=mergeSort(A,n,left,mid);
    v2=mergeSort(A,n,mid,right);
    v3=merge(A,n,left,mid,right);

    return v1+v2+v3;
  }
  else return 0;
}