luogu P1908 reverse order | Fenwick tree

Title Description

JERRY TOM cat and mouse contest recently, but after all, they are adults, they do not like to have to play with the kind of starting a game, and now they like to play statistics. Recently, TOM cat to a human Now called "reverse order" something, something which is defined as: For a given sequence of positive integers period, reverse and aj i <j is an ordered sequence of ai> Correct. Knowing this concept, whoever they match a given period is calculated positive integer number of sequences in the reverse order.
Update: Data has been strengthened.

Input Format

The first line, a number n, there is a sequence number n.

Second line number n, the given sequence. Each sequence number does not exceed 10 ^ 9

Output Format

A given number of the reverse sequence.

Description / Tips

For 25% of the data, n≤2500

For 50% of the data, n≤4 × 10 ^ 4

For all data, n≤5 × 10 ^ 5

Please use a faster input-output

N it should not be over 500,000 square bar by chen_zhe

---

#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
const int N=5e5+10;
#define int long long
using namespace std;
int c[N],n,a[N],b[N];
inline void add(int x,int y){
    for(;x<=n;x+=x&(-x))c[x]+=y;
}
inline int sum(int x){
    int ans=0;
    for(;x;x-=x&(-x))ans+=c[x];
    return ans;
}
signed main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        b[i]=a[i];
    }
    sort(b+1,b+1+n);
    int len=unique(b+1,b+1+n)-b-1;
    for(int i=1;i<=n;i++)
    a[i]=lower_bound(b+1,b+1+len,a[i])-b;
    int ans=0;
    for(int i=n;i>=1;i--){
        ans+=sum(a[i]-1);
        add(a[i],1);
    }
    cout<<ans<<endl;
}