牛客14522 珂朵莉的数列

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题解

统计逆序对的时候，乘以其所在的区间数目

另外，答案可能会超过long long存储范围，需要一点高精度

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct BIT
{
    ll bit[maxn], n;
    void init(int N){n=N;for(int i=1;i<=n;i++)bit[i]=0;}
    ll lowbit(ll x){return x&(-x);}
    void add(ll pos, ll v)
    {
        for(;pos<=n;pos+=lowbit(pos))bit[pos]+=v;
    }
    ll sum(ll pos)
    {
        ll ans(0);
        for(;pos;pos-=lowbit(pos))ans+=bit[pos];
        return ans;
    }
}bit;
struct Lisan
{
    int tmp[maxn], tot;
    void clear(){tot=0;}
    void insert(int x){tmp[++tot]=x;}
    void run()
    {
        sort(tmp+1,tmp+tot+1);
        tot=unique(tmp+1,tmp+tot+1)-tmp-1;
    }
    void lisan(ll *a, int len)
    {
        for(int i=1;i<=len;i++)a[i]=lower_bound(tmp+1,tmp+tot+1,a[i])-tmp;
    }
    int lisan(int x)
    {
        return lower_bound(tmp+1,tmp+tot+1,x)-tmp;
    }
}ls;
ll p[maxn], ans1=0, ans2=0;
int main()
{
    ll n, i, Base=1e15;
    n=read();
    ls.clear();
    rep(i,1,n)p[i]=read(), ls.insert(p[i]);
    ls.run();
    ls.lisan(p,n);
    bit.init(ls.tot);
    rep(i,1,n)
    {
        ll t = (n-i+1) * ( bit.sum(bit.n) - bit.sum(p[i]) );
        ans1 += t%Base;
        ans2 += t/Base;
        ans2 += ans1/Base;
        ans1 %= Base;
        bit.add(p[i],i);
    }
    if(ans2)printf("%lld%015lld",ans2,ans1);
    else printf("%lld",ans1);
    return 0;
}