結果

問題 No.1300 Sum of Inversions
ユーザー HyadoHyado
提出日時 2020-11-27 22:01:41
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 343 ms / 2,000 ms
コード長 5,364 bytes
コンパイル時間 1,999 ms
コンパイル使用メモリ 186,368 KB
実行使用メモリ 26,656 KB
最終ジャッジ日時 2024-07-26 12:45:16
合計ジャッジ時間 11,040 ms
ジャッジサーバーID
(参考情報)
judge5 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 34
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
template<typename T> using V = vector<T>;
template<typename T> using VV = vector<vector<T>>;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define all(v) (v).begin(),(v).end()
#define siz(v) (ll)(v).size()
#define rep(i,a,n) for(ll i=a;i<(ll)(n);++i)
#define repr(i,a,n) for(ll i=n-1;(ll)a<=i;--i)
#define ENDL '\n'
typedef pair<int,int> Pi;
typedef pair<ll,ll> PL;
constexpr ll mod = 998244353;
constexpr ll INF = 1000000099;
constexpr ll LINF = (ll)(1e18 +99);
const ld PI = acos((ld)-1);
const vector<ll> dx={-1,0,1,0},dy={0,1,0,-1};
template<typename T,typename U> inline bool chmin(T& t, const U& u){if(t>u){t=u;return 1;}return 0;}
template<typename T,typename U> inline bool chmax(T& t, const U& u){if(t<u){t=u;return 1;}return 0;}
template<typename T> inline T gcd(T a,T b){return b?gcd(b,a%b):a;}
inline void yes() { cout << "Yes" << ENDL; }
inline void no() { cout << "No" << ENDL; }
template<typename T,typename Y> inline T mpow(T a, Y n) {
T res = 1;
for(;n;n>>=1) {
if (n & 1) res = res * a;
a = a * a;
}
return res;
}
template <typename T> V<T> prefix_sum(const V<T>& v) {
int n = v.size();
V<T> ret(n + 1);
rep(i, 0, n) ret[i + 1] = ret[i] + v[i];
return ret;
}
template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
for(auto&& x: vec) is >> x;
return is;
}
template<typename T,typename Y>
ostream& operator<<(ostream& os,const pair<T,Y>& p){
return os<<"{"<<p.fs<<","<<p.sc<<"}";
}
template<typename T> ostream& operator<<(ostream& os,const V<T>& v){
os<<"{";
for(auto e:v)os<<e<<",";
return os<<"}";
}
template<typename ...Args>
void debug(Args&... args){
for(auto const& x:{args...}){
cerr<<x<<' ';
}
cerr<<ENDL;
}
template <long long MOD = mod> // MOD
struct Mint {
using M = Mint;
ll a;
Mint(ll x = 0) {
a = x % MOD;
if(a < 0) a += MOD;
} // Mint
M pow(ll n) {
M b = *this, r = 1 % MOD;
while(n) {
if(n & 1) r *= b;
b *= b;
n >>= 1;
}
return r;
}
M inv() { return pow(MOD - 2); }
M& operator+=(M r) {
a += r.a;
if(a >= MOD) a -= MOD;
return *this;
}
M& operator-=(M r) {
a += MOD - r.a;
if(a >= MOD) a -= MOD;
return *this;
}
M& operator*=(M r) {
a = 1LL * a * r.a % MOD;
return *this;
}
M& operator/=(M r) { return (*this) *= r.inv(); }
M operator+(M r) const { return M(a) += r; };
M operator-(M r) const { return M(a) -= r; };
M operator*(M r) const { return M(a) *= r; };
M operator/(M r) const { return M(a) /= r; };
M operator-() const { return M() - *this; }
friend ostream& operator<<(ostream& os, const M& r) { return os << r.a; }
};
using M = Mint<mod>;
/*
using M = Mint<mod>;
M xint ix.a+=i x+=M{i}
M(x).pow(exp)
*/
template <typename T> struct BIT {
int n;
vector<T> bit; // 1-indexed
BIT() : n(-1) {}
BIT(int n_, T d) : n(n_), bit(n_ + 1, d) {}
// initialization2 n_ d
T sum(int i) {
T s = bit[0];
for(int x = i + 1; x > 0; x -= (x & -x)) s += bit[x];
return s;
}
// 1i(1-indexed)
void add(int i, T a) {
for(int x = i + 1; x <= n; x += (x & -x)) bit[x] += a;
}
// ia
T lower_bound(T w) {
if(w <= 0) return 0;
T x = 0, r = 1; // x
while(r < n) r <<= 1;
for(T k = r; k > 0; k >>= 1) { //
if(x + k <= n && bit[x + k] < w) {
w -= bit[x + k];
x += k; //
}
}
return x + 1;
}
// indwind
T query(int l, int r) { return sum(r - 1) - sum(l - 1); } // [l,r)(0-indexed)
};
template<typename T>
struct compression{
V<T> tmp;
map<T,T> ma;
void push(T x){
tmp.emplace_back(x);
}
void build(){
sort(all(tmp));
tmp.erase(unique(all(tmp)),tmp.end());
rep(i,0,siz(tmp))ma[tmp[i]]=i;//
}
void change(T& x){
assert(ma.count(x));
x=ma[x];
}
void change(V<T>& v){
rep(i,0,siz(v)){
assert(ma.count(v[i]));
v[i]=ma[v[i]];
}
}
};
signed main(){
cin.tie(0);cerr.tie(0);ios::sync_with_stdio(false);
cout<<fixed<<setprecision(20);
ll n;cin>>n;
V<ll> v(n);cin>>v;
V<ll> id=v;
compression<ll> com;
rep(i,0,n)com.push(v[i]);
com.build();
com.change(id);
BIT<ll> bl(n+5,0),br(n+5,0),numl(n+5,0),numr(n+5,0);
rep(i,0,n){
br.add(id[i],v[i]);
numr.add(id[i],1);
}
M ans=0;
rep(j,0,n){
br.add(id[j],-v[j]);
numr.add(id[j],-1);
M l=numl.query(id[j]+1,n),r=numr.query(0,id[j]);
ans+=r*bl.query(id[j]+1,n)+l*br.query(0,id[j])+r*l*v[j];
bl.add(id[j],v[j]);
numl.add(id[j],1);
}
cout<<ans<<ENDL;
}
//! ( . _ . ) !
//CHECK overflow,vector_size,what to output?
//any other simpler approach?
//list all conditions, try mathematical and graphic observation
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