답안 #968241

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
968241 2024-04-23T08:39:18 Z gaga999 Real Mountains (CCO23_day1problem2) C++17
0 / 25
6 ms 27124 KB
// #pragma GCC optimize("Ofast,no-stack-protector")
// #pragma GCC optimize("O3,unroll-loops")
// #pragma GCC target("avx,avx2,bmi,bmi2,lzcnt,popcnt")
#include <bits/stdc++.h>
#define lowbit(x) ((x) & -(x))
#define ml(a, b) ((1ll * (a) * (b)) % M)
#define tml(a, b) (a) = ((1ll * (a) * (b)) % M)
#define ad(a, b) ((0ll + (a) + (b)) % M)
#define tad(a, b) (a) = ((0ll + (a) + (b)) % M)
#define mi(a, b) ((0ll + M + (a) - (b)) % M)
#define tmi(a, b) (a) = ((0ll + M + (a) - (b)) % M)
#define tmin(a, b) (a) = min((a), (b))
#define tmax(a, b) (a) = max((a), (b))
#define iter(a) (a).begin(), (a).end()
#define riter(a) (a).rbegin(), (a).rend()
#define init(a, b) memset((a), (b), sizeof(a))
#define cpy(a, b) memcpy((a), (b), sizeof(a))
#define uni(a) a.resize(unique(iter(a)) - a.begin())
#define size(x) (int)x.size()
#define pb emplace_back
#define mpr make_pair
#define ls(i) ((i) << 1)
#define rs(i) ((i) << 1 | 1)
#define INF 0x3f3f3f3f
#define NIF 0xc0c0c0c0
#define eps 1e-9
#define F first
#define S second
#define AC cin.tie(0)->sync_with_stdio(0)
using namespace std;
typedef long long llt;
typedef __int128_t lll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<llt, llt> pll;
typedef complex<double> cd;
const int M = 1e6 + 3;

// random_device rm;
// mt19937 rg(rm());
// default_random_engine rg(rm());
// uniform_int_distribution<int> rd(INT_MIN, INT_MAX);
// uniform_real_distribution<double> rd(0, M_PI);

void db() { cerr << "\n"; }
template <class T, class... U>
void db(T a, U... b) { cerr << a << " ", db(b...); }

inline char gc()
{
    const static int SZ = 1 << 16;
    static char buf[SZ], *p1, *p2;
    if (p1 == p2 && (p2 = buf + fread(p1 = buf, 1, SZ, stdin), p1 == p2))
        return -1;
    return *p1++;
}
void rd() {}
template <typename T, typename... U>
void rd(T &x, U &...y)
{
    x = 0;
    bool f = 0;
    char c = gc();
    while (!isdigit(c))
        f ^= !(c ^ 45), c = gc();
    while (isdigit(c))
        x = (x << 1) + (x << 3) + (c ^ 48), c = gc();
    f && (x = -x), rd(y...);
}

template <typename T>
void prt(T x)
{
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        prt(x / 10);
    putchar((x % 10) ^ 48);
}

const int N = 1e6 + 6;
int h[N], tr[N << 1], ln;
vector<int> po;
vector<int> pl[N];
bool vd[N];

inline int fd(int x)
{
    return lower_bound(iter(po), x) - po.begin();
}

signed main()
{
    int n;
    rd(n);
    for (int i = 1; i <= n; i++)
        rd(h[i]), tr[i + n - 1] = h[i];
#define pul(x) tr[x] = min(tr[ls(x)], tr[rs(x)])
    for (int i = n - 1; i > 0; i--)
        pul(i);
    auto cg = [&](int p, int v) -> void
    {
        p += n - 1, tr[p] = v;
        for (p >>= 1; p; p >>= 1)
            pul(p);
    };
    auto qy = [&](int l, int r) -> int
    {
        int mn = INF;
        for (l += n - 1, r += n - 1; l <= r; l >>= 1, r >>= 1)
        {
            if (l & 1)
                tmin(mn, tr[l++]);
            if (~r & 1)
                tmin(mn, tr[r--]);
        }
        return mn;
    };
    po = vector<int>(h + 1, h + n + 1);
    sort(iter(po)), uni(po);
    ln = size(po);
    for (int i = 1; i <= n; i++)
        h[i] = fd(h[i]), pl[h[i]].pb(i);
    int vl = 1, vr = n;
    set<int> st;
    llt ans = 0;
    for (int i = 0; i < ln; i++)
    {
        for (int j : pl[i])
            vd[j] = 1, st.insert(j), cg(j, INF);
        while (vd[vl])
            vl++;
        while (vd[vr])
            vr--;
        while (!st.empty() && vl >= *st.begin())
            st.erase(st.begin());
        while (!st.empty() && vr <= *st.rbegin())
            st.erase(prev(st.end()));
        if (st.empty())
            continue;
        int num = size(st), d = po[i + 1] - po[i];
        llt sum = ml((po[i + 1] - 1ll + po[i]) >> 1, d);
        sum %= M;
        int v0 = qy(1, *st.begin()), v2 = qy(*st.rbegin(), n);
        if (num == 1)
        {
            ans += (v0 + v2) * d + sum;
        }
        else
        {
            ans += sum * num;
            sum += po[i + 1] - po[i];
            ans += sum * ((num << 1) - 3ll);
            int v1 = qy(*st.begin(), *st.rbegin());
            ans += (0ll + v0 + v2 + min({v0, v1, v2})) * d;
        }
        ans %= M;
    }
    prt(ans % M), putchar('\n');
}
# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 26972 KB Output is correct
2 Incorrect 6 ms 27124 KB Output isn't correct
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 26972 KB Output is correct
2 Incorrect 6 ms 27124 KB Output isn't correct
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 26972 KB Output is correct
2 Incorrect 6 ms 27124 KB Output isn't correct
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 26972 KB Output is correct
2 Incorrect 6 ms 27124 KB Output isn't correct
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 26972 KB Output is correct
2 Incorrect 6 ms 27124 KB Output isn't correct
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 26972 KB Output is correct
2 Incorrect 6 ms 27124 KB Output isn't correct
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 26972 KB Output is correct
2 Incorrect 6 ms 27124 KB Output isn't correct
3 Halted 0 ms 0 KB -