acwing 788. Number of pairs in reverse order

subject ：788. The number of reverse order pairs

Ideas ： Sort by merging , Then when a[i[>a[j] when ,ct+=mid-i+1 that will do , Pay attention to longlong

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;
typed![ Insert picture description here ](https://img-blog.csdnimg.cn/f0c8e2c9eed3414eb39d822e812e62ee.png?x-oss-process=image/watermark,type_d3F5LXplbmhlaQ,shadow_50,text_Q1NETiBA5bKB5b-n,size_20,color_FFFFFF,t_70,g_se,x_16)
ef long long LL;
int n,a[100010];

LL merge(int l,int r){
    
    if(l>=r) return 0;
    int mid=l+r>>1;
    LL ct=0;
    ct+=merge(l,mid);
    ct+=merge(mid+1,r);
    int t[100010],k=0;
    int i=l,j=mid+1;
    
    while(i<=mid&&j<=r){
    
        if(a[i]<=a[j]) t[k++]=a[i++];
        else {
    
            ct+=mid-i+1;
            t[k++]=a[j++];
        }
        
    }
    while(i<=mid) t[k++]=a[i++];
    while(j<=r) t[k++]=a[j++];
    for(int i=0;i<k;i++){
    
        a[i+l]=t[i];
    }
    return ct;
}
    
int main(){
    
    scanf("%d",&n);
    for(int i=0;i<n;i++){
    
        scanf("%d",&a[i]);
    }
    printf("%lld",merge(0,n-1));
    return 0;
}