Submission #1049988

#	Time	Username	Problem	Language	Result	Execution time	Memory
1049988		shiomusubi496	Uplifting Excursion (BOI22_vault)	C++17	30 / 100	5065 ms	24052 KiB

vault

#include <bits/stdc++.h>

#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rep2(i, a, b) for (int i = (int)(a); i < (int)(b); ++i)
#define rrep(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define rrep2(i, a, b) for (int i = (int)(b) - 1; i >= (int)(a); --i)
#define all(v) begin(v), end(v)
#define rall(v) rbegin(v), rend(v)

using namespace std;

using ll = long long;

template<class T, class U> bool chmin(T& a, const U& b) { return a > b ? a = b, true : false; }
template<class T, class U> bool chmax(T& a, const U& b) { return a < b ? a = b, true : false; }

using namespace std;

constexpr ll inf = 1e18;

ll gcd(ll a, ll b) {
    a = abs(a); b = abs(b);
    while (b) {
        a %= b;
        swap(a, b);
    }
    return a;
}

int main() {
    int M; cin >> M;
    ll L; cin >> L;
    vector<ll> A(2 * M + 1);
    rep (i, 2 * M + 1) cin >> A[i];
    if (L < 0) {
        reverse(all(A));
        L = -L;
    }
    ll sml = 0, smr = 0;
    rep (i, 2 * M + 1) {
        if (i < M) sml += A[i] * (i - M);
        if (i > M) smr += A[i] * (i - M);
    }
    if (L < sml || smr < L) {
        puts("impossible");
        return 0;
    }
    vector<pair<int, ll>> B;
    vector<pair<int, ll>> C;
    rep (i, 2 * M + 1) {
        if (i == M) continue;
        if (A[i] == 0) continue;
        if (A[i] <= M) B.emplace_back(i - M, A[i]);
        else C.emplace_back(i - M, A[i]);
    }
    sml = smr = 0;
    for (auto [x, y] : B) {
        if (x < 0) sml += x * y;
        else smr += x * y;
    }
    auto calc_dp = [](vector<pair<int, ll>> B) -> vector<ll> {
        ll sml = 0, smr = 0;
        for (auto [x, y] : B) {
            if (x < 0) sml += x * y;
            else smr += x * y;
        }
        vector<ll> dp(smr - sml + 1, -inf);
        dp[0 - sml] = 0;
        for (auto [x, y] : B) {
            int ax = abs(x);
            vector<ll> memo;
            vector<ll> res;
            deque<int> st;
            rep (i, ax) {
                // mod x ごとにスライド最大値の形になる
                memo.clear();
                for (int j = i; j <= smr - sml; j += ax) {
                    memo.push_back(dp[j]);
                }
                res.assign(memo.size(), -inf);
                st.clear();
                if (x > 0) {
                    rep (i, memo.size()) memo[i] -= i;
                    rep (i, res.size()) {
                        while (!st.empty() && st.front() < i - y) st.pop_front();
                        while (!st.empty() && memo[st.back()] < memo[i]) st.pop_back();
                        st.push_back(i);
                        res[i] = memo[st.front()] + i;
                    }
                }
                else {
                    rep (i, memo.size()) memo[i] += i;
                    rrep (i, res.size()) {
                        while (!st.empty() && st.front() > i + y) st.pop_front();
                        while (!st.empty() && memo[st.back()] < memo[i]) st.pop_back();
                        st.push_back(i);
                        res[i] = memo[st.front()] - i;
                    }
                }
                rep (j, res.size()) dp[j * ax + i] = res[j];
            }
        }
        return dp;
    };
    auto dp = calc_dp(B);
    if (C.empty()) {
        if (dp[L - sml] >= 0) {
            cout << dp[L - sml] + A[M] << endl;
        }
        else {
            puts("impossible");
        }
        return 0;
    }
    if (C.size() <= 1) {
        auto [x, y] = C[0];
        ll lo = 0, hi = x * y;
        if (lo > hi) swap(lo, hi);
        ll ans = -inf;
        rep (i, smr - sml + 1) {
            if (dp[i] < 0) continue;
            if (L < (i + sml) + lo || (i + sml) + hi < L) continue;
            if ((L - (i + sml)) % x) continue;
            chmax(ans, dp[i] + A[M] + abs((L - (i + sml)) / x));
        }
        if (ans >= 0) {
            cout << ans << endl;
        }
        else {
            puts("impossible");
        }
        return 0;
    }
    ll ans = -inf;
    vector<ll> g(C.size() + 1);
    rrep (i, C.size()) g[i] = gcd(g[i + 1], C[i].first);
    rep (i, smr - sml + 1) {
        if (dp[i] < 0) continue;
        ll L2 = L - (i + sml);
        ll sm = dp[i] + A[M];
        int m = -1;
        rep (j, C.size() - 1) {
            ll tmp = C[j].second;
            while ((L2 - tmp * C[j].first) % g[j + 1]) --tmp;
            assert(tmp >= 0);
            if ((L2 - tmp * C[j].first) / g[j + 1] < M * M) {
                m = j;
                break;
            }
            sm += tmp;
            L2 -= tmp * C[j].first;
        }
        if (m == -1) {
            assert(L2 % C.back().first == 0);
            ll tmp = L2 / C.back().first;
            if (tmp > C.back().second) continue;
            sm += tmp;
            chmax(ans, sm);
        }
        else {
            vector<pair<int, ll>> D;
            rep2 (j, m + 1, C.size()) D.emplace_back(C[j].first, min<ll>(C[j].second, 2 * M));
            auto dp2 = calc_dp(D);
            ll sml2 = 0, smr2 = 0;
            for (auto [x, y] : D) {
                if (x < 0) sml2 += x * y;
                else smr2 += x * y;
            }
            rep (j, smr2 - sml2 + 1) {
                if (dp2[j] < 0) continue;
                ll L3 = L2 - (j + sml2);
                ll sm2 = sm + dp2[j];
                if (L3 % C[m].first) continue;
                ll tmp = L3 / C[m].first;
                if (tmp > C[m].second) continue;
                if (tmp < 0) continue;
                sm2 += tmp;
                chmax(ans, sm2);
            }
        }
    }
    if (ans >= 0) {
        cout << ans << endl;
    }
    else {
        puts("impossible");
    }
}

/*
フロベニウスの硬貨問題 : 互いに素な a,b で作れない最大の値は ab-a-b
x 以下を全て使う払い方があれば、それを必ず採用するとして良い (x 以下を使わない代わりに x 以上を使うのは明らかに少なくなる)
の 2 つから、 L-M^2 を超えない範囲で小さい方から選べるだけ選ぶ戦略が取れる
互いに素じゃない場合が面倒くさい
*/

• Subtask 1: 5 / 5
• Subtask 2: 15 / 15
• Subtask 3-1: 10 / 10
• Subtask 3-2: 0 / 10
• Subtask 4-1: 0 / 10
• Subtask 4-2: 0 / 10
• Subtask 5-1: 0 / 10
• Subtask 5-2: 0 / 10
• Subtask 6-1: 0 / 10
• Subtask 6-2: 0 / 10