結果

問題 No.3129 Multiple of Twin Subarray
ユーザー hint908
提出日時 2025-04-25 21:42:32
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 311 ms / 2,000 ms
コード長 3,998 bytes
コンパイル時間 3,905 ms
コンパイル使用メモリ 287,288 KB
実行使用メモリ 51,828 KB
最終ジャッジ日時 2025-04-25 21:42:51
合計ジャッジ時間 13,991 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
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 * 46
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 # 

// #pragma GCC target("avx2")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")


#include<bits/stdc++.h>
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template<class T> using V = vector<T>;
template<class T> using VV = V<V<T>>;
template<class T> using VVV = V<VV<T>>;
template<class T> using VVVV = VV<VV<T>>;
#define rep(i,n) for(ll i=0ll;(i)<(n);(i)++)
#define REP(i,a,n) for(ll i=(a);(i)<(n);(i)++)
#define rrep(i,n) for(ll i=(n)-1;(i)>=(0ll);(i)--)
#define RREP(i,a,n) for(ll i=(n)-1;(i)>=(a);(i)--)
const long long INF = (1LL << 60);
const long long mod99 = 998244353;
const long long mod107 = 1000000007;
const long long mod = mod99;
#define eb emplace_back
#define be(v) (v).begin(),(v).end()
#define all(v) (v).begin(),(v).end()
#define foa(i,v) for(auto& (i) : (v))
#define UQ(v) sort(be(v)), (v).erase(unique(be(v)), (v).end())
#define UQ2(v,cmp) sort(be(v)), (v).erase(unique(be(v),cmp), (v).end())
#define UQ3(v,cmp) sort(be(v),cmp), (v).erase(unique(be(v)), (v).end())
#define UQ4(v,cmp,cmp2) sort(be(v), cmp), (v).erase(unique(be(v),cmp2), (v).end())
#define LB(x,v) (lower_bound(be(v),(x))-(v).begin())
#define LB2(x,v,cmp) (lower_bound(be(v),(x),(cmp))-(v).begin())
#define UB(x,v) (upper_bound(be(v),(x))-(v).begin())
#define UB2(x,v,cmp) (upper_bound(be(v),(x),(cmp))-(v).begin())
#define dout()  cout << fixed << setprecision(20)
#define randinit() srand((unsigned)time(NULL))

template<class T, class U> bool chmin(T& t, const U& u) { if (t > u){ t = u; return 1;} return 0; }
template<class T, class U> bool chmax(T& t, const U& u) { if (t < u){ t = u; return 1;} return 0; }


ll Rnd(ll L=0, ll R=mod99){return rand()%(R-L)+L;}

template <typename T, T (*OP)(T, T), T (*E)()>
struct SegmentTree { 
  vector<T> seg;
  ll seg_size;//葉の数
    
  SegmentTree(ll N) : seg(4*N, E()), seg_size() {
    seg_size = 1;
    while(seg_size < N) seg_size *= 2;
  }
  SegmentTree(const vector<T> a) : seg(4*a.size(), E()), seg_size() {
    ll N = a.size();
    seg_size = 1;
    while(seg_size < N) seg_size *= 2;
    rep(i,N) seg[i + seg_size - 1] = a[i];
    for(ll i = seg_size-2; i >= 0; i --){
      seg[i] = OP(seg[2*i + 1], seg[2*i + 2]);
    }
  }
  
  void set(ll idx, T x){
    idx += seg_size - 1;
    seg[idx] = x;
    
    while(idx > 0){
      idx = (idx - 1) / 2;
      seg[idx] = OP(seg[2*idx + 1], seg[2*idx + 2]);
    }
  }

  void apply(ll idx, T x){
    idx += seg_size - 1;
    seg[idx] = OP(seg[idx], x);
    
    while(idx > 0){
      idx = (idx - 1) / 2;
      seg[idx] = OP(seg[2*idx + 1], seg[2*idx + 2]);
    }
  }
 

  T prod(ll a, ll b) { return prod_sub(a, b, 0, 0, seg_size); }
  T prod_sub(ll a, ll b, ll idx, ll left, ll right){
    if(a >= right || b <= left) return E();
    if(a <= left && b >= right) return seg[idx];
  
    T v1 = prod_sub(a, b, 2 * idx + 1, left, (left + right) / 2);
    T v2 = prod_sub(a, b, 2 * idx + 2, (left + right) / 2, right);
    return OP(v1, v2);
  }
  
};

ll e(){return -INF;}
ll op(ll L, ll R){return max(L,R);}



ll e2(){return INF;}
ll op2(ll L, ll R){return min(L,R);}



void solve(){
    ll n;
    cin >> n;
    V<ll> v(n);
    rep(i,n) cin >> v[i];
    V<ll> sv(n+1, 0);
    rep(i,n) sv[i+1] = sv[i] + v[i];
    SegmentTree<ll, op, e> seg(sv);
    SegmentTree<ll, op2, e2> seg2(sv);
    V<ll> a(n), b(n), c(n), d(n);
    rep(i, n){
        a[i] = sv[i+1] - seg.prod(0, i+1);
        b[i] = sv[i+1] - seg2.prod(0, i+1);
        c[i] = seg.prod(i+1, n+1) - sv[i];
        d[i] = seg2.prod(i+1, n+1) - sv[i];
    }
    
    SegmentTree<ll, op, e> B(b), C(c);
    SegmentTree<ll, op2, e2> A(a), D(d);
    ll ans = -INF;
    rep(i, n-1){
        chmax(ans, B.prod(0, i+1) * C.prod(i+1, n));
        chmax(ans, A.prod(0, i+1) * D.prod(i+1, n));
    }
    cout << ans << endl;
    
    
}





int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int t=1;
    // cin >> t;
    rep(i,t) solve();
}
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