C++ number of reverse pairs (merge sort)

Given an integer sequence of length n, please count the number of reverse pairs in the sequence.
The definition of the reverse order pair is as follows: For the i-th and j-th elements of the sequence, if i <j and a[i]> a[j] are satisfied, then it is a reverse-order pair; otherwise, it is not.
Input format The
first line contains the integer n, which represents the length of the sequence.
The second line contains n integers, representing the entire sequence.
Output format
Output an integer, indicating the number of pairs in reverse order.
Data range
1≤n≤100000
Input example:
6
2 3 4 5 6 1
Output example:
5

Merge sorting is a powerful tool for finding reverse order pairs, and the monotonicity brought by sorting is used to find reverse order pairs at a time.

AC code:

#include<stdio.h>

int n;
int a[100010];
int temp[100010];
long long ans;

void merge_sort(int l,int r)
{
    
    
    if(l==r) return;

    int mid=l+((r-l)>>1);
    merge_sort(l,mid);
    merge_sort(mid+1,r);

    for(int i=l;i<=r;++i) temp[i]=a[i];

    int i1=l,i2=mid+1;
    for(int i=l;i<=r;++i)
    {
    
    
        if(i1>mid) a[i]=temp[i2++];
        else if(i2>r)  a[i]=temp[i1++];
        else if(temp[i1]<=temp[i2]) a[i]=temp[i1++];
        //左半边i1之后的元素都比i2大
        else if(temp[i1]>temp[i2]){
    
    ans+=mid-i1+1;a[i]=temp[i2++];}
    }
}

int main()
{
    
    
    scanf("%d",&n);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]);
    merge_sort(1,n);
    printf("%lld",ans);
    return 0;
}