#1154640 (C++23) No.1300 Sum of Inversions

• 問題: No.1300 Sum of Inversions
• ユーザー: addx
• 提出日時: 2026-03-31 15:23:12
• 言語: C++23 (gcc 15.2.0 + boost 1.89.0)
• 結果: WA
• 実行時間: -
• コード長: 3,393 bytes
• コンパイル時間: 4,597 ms
• コンパイル使用メモリ: 384,432 KB
• 実行使用メモリ: 404,760 KB
• 最終ジャッジ日時: 2026-03-31 15:23:36
• 合計ジャッジ時間: 10,550 ms
• ジャッジサーバーID （参考情報）: judge3_1 / judge2_1

ファイルパターン	結果
sample	AC * 3
other	AC * 2 WA * 32

ソースコード

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
//using namespace atcoder;
using ll = long long;
using ull = unsigned long long;
using i128 = __int128_t;
using u128 = unsigned __int128_t;
using mint = atcoder::static_modint<998244353>;
const int mod = 998244353;
#include <chrono>
mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());
int dx[8] = {-1, 1, 0, 0, -1, -1, 1, 1};
int dy[8] = {0, 0, -1, 1, -1, 1, -1, 1};
template<class T>
bool chmin(T& a, const T& b){
    if (b < a){
        a = b;
        return true;
    }
    else { 
        return false;
    }
}
template<class T> 
bool chmax(T& a, const T& b){
    if (a < b){
        a = b;
        return true;
    }
    else {
        return false;
    }
}
namespace addx{

    template <typename T>
    class FenwickTree{
    private:
        int n;
        vector<T> t;
    public:
        FenwickTree(int size) : n(size), t(size + 1, T(0)) {}
        //O(n)構築
        FenwickTree(const vector<T>& x){
            n = (int)x.size();
            t.assign(n + 1, T(0));
            for (int i = 1; i <= n; i++){
                t[i] += x[i - 1];
                int k = i + (i & -i);
                if (k <= n){
                    t[k] += t[i];
                }
            }
        }
        void add(int idx, T val){
            for (int i = idx + 1; i <= n; i += i & -i){
                t[i] += val;
            }
        }
        T sum(int idx){
            T now = 0;
            for (int i = idx; i > 0; i -= i & -i){
                now += t[i];
            }
            return now;
        }
        T sum(int l, int r){
            return sum(r) - sum(l);
        }
        //初めてsum >= kになるindexを返す
        int max_right(T k){
            if (k <= 0){
                return 0;
            }
            int idx = 0;
            int bt = 1;
            while((bt << 1) <= n){
                bt <<= 1;
            }
            T sm = 0;
            for (int i = bt; i > 0; i >>= 1){
                int nxt = idx + i;
                if (nxt <= n && sm + t[nxt] < k){
                    idx = nxt;
                    sm += t[nxt];
                } 
            }
            return idx;
        }    
    };
}
void solve(){
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++){
        cin >> a[i];
    }
    vector<int> b = a;
    sort(b.begin(), b.end());
    b.erase(unique(b.begin(), b.end()), b.end());
    unordered_map<int, int> mp;
    for (int i = 0; i < (int)b.size(); i++){
        mp[b[i]] = i;
    }
    addx::FenwickTree<ll> fw1((int)b.size()), fw2((int)b.size()), fw3((int)b.size()), fw4((int)b.size());
    for (int i = 0; i < n; i++){
        fw1.add(mp[a[i]], a[i]);
        fw2.add(mp[a[i]], 1);
    }
    ll ans = 0;
    for (int i = 0; i < n; i++){
        int idx = mp[a[i]];
        fw1.add(idx, -a[i]);
        fw2.add(idx, -1);
        ll l = fw2.sum(0, idx), r = fw4.sum(idx + 1, (int)b.size());
        if (l != 0 && r != 0){
            ans += l * r * a[i] + fw1.sum(0, idx) * r + fw3.sum(idx + 1, (int)b.size()) * l;
        }
        fw3.add(idx, a[i]);
        fw4.add(idx, 1);
    }
    cout << ans << '\n';
};

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;   
    while(t--){
        cout << fixed << setprecision(15);          
        solve();
    }   
}