珂朵莉的數列（樹狀數組）

思路:對於a[L]>a[R] L<R 逆序數對所被包含的區間是左端點(1——L)到右端點（R——n）所以他的貢獻是L*(n-r+1)我們用樹狀數組去求逆序數對，答案會爆ll用int128

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e4+7;
const int N=1e6+10;
const int inf=0x7f7f7f7f;


ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
    return a*(b/gcd(a,b));
}

template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-')
            op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op)
        x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0)
        x = -x, putchar('-');
    if(x >= 10)
        write(x / 10);
    putchar('0' + x % 10);
}

ll a[N],b[N];
vector<int> vt;
int getid(int x)
{
    return lower_bound(vt.begin(),vt.end(),x)-vt.begin()+1;
}
int lowbit(int x)
{
    return x&-x;
}
void add(int x,int k)
{

    while(x<N)
    {
        b[x]+=k;
        x+=lowbit(x);
    }

}
ll sum(int x)
{
    ll ans=0;
    while(x>0)
    {
        ans+=b[x];
        x-=lowbit(x);
    }
    return ans;
}
void print(__int128 x)
{
    if(x<10)
    {
        putchar(x%10+'0');
        return ;
    }
    print(x/10);
     putchar(x%10+'0');
}


int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        vt.push_back(a[i]);
    }
    sort(vt.begin(),vt.end());
    vt.erase(unique(vt.begin(),vt.end()),vt.end());
    __int128 ans=0;
    for(ll i=1;i<=n;i++)
    {
        int pos=getid(a[i]);
        ans+=(ll)(sum(N-1)-sum(pos))*(n-i+1);
        add(pos,i);
    }
    print(ans);



    return 0;
}