Acwing 788.逆序对的数量

#include<bits/stdc++.h>
using namespace std;
 
const int N=1e+5;
int  a[N],tmp[N];
typedef long long ll;  #注意题目条件
 
ll merge_sort(int q[],int l,int r){
    if(l>=r) return 0;
    int mid=l+r>>1;
    
    ll res=merge_sort(q,l,mid)+merge_sort(q,mid+1,r);  #注意
	 
    int k=0,i=l,j=mid+1;
    while(i<=mid&&j<=r){
    	if(q[i]<=q[j])
    		tmp[k++]=q[i++];
    	else{
   			res+=mid-i+1;  #回忆计算过程（力扣视频上有）
   			tmp[k++]=q[j++];
		}	
	}
	
	while(i<=mid)
		tmp[k++]=q[i++];
	while(j<=r)
		tmp[k++]=q[j++];
		
	for(i=l,j=0;i<=r;i++,j++)
		q[i]=tmp[j];
	
	return res;
 
}
 
int main(){
    int n;
    cin>>n;
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
    cout<<merge_sort(a,0,n-1)<<endl;
    return 0;
}