逆序对的数量

#include<iostream>
using namespace std;
int num[100010], temp[100010];
long long count( int l , int r){
    if(l >= r) return 0;
    
    long long sum = 0;
    int mid = (l + r) >> 1;
    
    sum += count(l , mid);
    sum += count(mid + 1 , r);
    
    for(int i = l , j = mid + 1; i <= mid && j <= r; ){
        if(num[i] > num[j]) {
            sum += mid - i + 1;
            j ++;
        }
        else i ++;
    }
    
    int i = l , j = mid + 1 , k = 0;
    while(i <= mid && j <= r){
        if(num[i] <= num[j]) temp[k ++] = num[i ++];
        else temp[k ++] = num[j ++]; 
    }
    while(i <= mid) temp[k ++] = num[i ++];
    while(j <= r) temp[k ++] = num[j ++];
    
    for(int i = l , j = 0 ; j < k ; j ++ , i ++) num[i] = temp[j];
    
    return sum;
}

int main(){
    int n ; 
    cin >> n;
    
    for(int i = 0 ; i < n ; i ++) cin >> num[i];
    
    cout << count(0 , n - 1) << endl;
    
    return 0;
}