結果

問題 No.2860 Heal Slimes
ユーザー dyktr_06
提出日時 2025-04-18 16:05:38
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,848 bytes
コンパイル時間 2,081 ms
コンパイル使用メモリ 195,944 KB
実行使用メモリ 7,848 KB
最終ジャッジ日時 2025-04-18 16:05:48
合計ジャッジ時間 7,251 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
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ファイルパターン 結果
other WA * 20 RE * 40
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ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

// o4-mini-high
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, K;
    ll X;
    cin >> N >> K >> X;
    vector<ll> H(N);
    for (int i = 0; i < N; i++) {
        cin >> H[i];
    }

    // If K == N, we can only add to all N slimes at once:
    // differences remain the same, so only possible if all H are already equal.
    if (K == N) {
        bool all_eq = true;
        for (int i = 1; i < N; i++) {
            if (H[i] != H[0]) {
                all_eq = false;
                break;
            }
        }
        cout << (all_eq ? "Yes\n" : "No\n");
        return 0;
    }

    // Check the modulo invariant: all H[i] % X must be the same.
    ll r = H[0] % X;
    for (int i = 1; i < N; i++) {
        if (H[i] % X != r) {
            cout << "No\n";
            return 0;
        }
    }

    // Build a[i] = (H[i] - r) / X, so H[i] = X * a[i] + r.
    vector<ll> a(N+1);
    for (int i = 1; i <= N; i++) {
        a[i] = (H[i-1] - r) / X;
    }

    // Build the "delta" array d[i] = a[i] - a[i-1] for i=2..N, and d[N+1] = -a[N].
    vector<ll> d(N+2);
    for (int i = 2; i <= N; i++) {
        d[i] = a[i] - a[i-1];
    }
    d[N+1] = -a[N];

    int M = N - K + 1;  // number of possible operation start positions

    // Compute base_op[j] for j=1..M, assuming the final target a_f = 0.
    // Later we'll adjust by a global C = a_f (the target a-value).
    vector<ll> base_op(M+1);
    base_op[1] = -a[1];
    for (int j = 2; j <= M; j++) {
        if (j <= K) {
            // op_j = -d[j]
            base_op[j] = -d[j];
        } else {
            // op_j = -d[j] + op_{j-K}
            base_op[j] = -d[j] + base_op[j - K];
        }
    }

    // For starts j where (j-1)%K != 0, base_op[j] must be >= 0 (no C adjustment).
    for (int j = 1; j <= M; j++) {
        if ((j - 1) % K != 0) {
            if (base_op[j] < 0) {
                cout << "No\n";
                return 0;
            }
        }
    }

    // Compute the minimum C needed to make op_j >= 0 for j ≡ 1 (mod K):
    // op_j = base_op[j] + C >= 0  =>  C >= -base_op[j]
    ll C_min = LLONG_MIN;
    for (int j = 1; j <= M; j++) {
        if ((j - 1) % K == 0) {
            C_min = max(C_min, -base_op[j]);
        }
    }

    // Tail consistency constraints:
    // For i = M+1..N, let j = i-K (so j in [N-2K+2 .. N-K]),
    // we need d[i] = op_j.
    int j_low = max(1, N - 2 * K + 2);
    int j_high = N - K;

    bool C_assigned = false;
    ll C_val = 0;

    for (int j = j_low; j <= j_high; j++) {
        int i = j + K;  // the corresponding delta index
        if ((j - 1) % K == 0) {
            // op_j = base_op[j] + C must equal d[i]
            ll neededC = d[i] - base_op[j];
            if (!C_assigned) {
                C_assigned = true;
                C_val = neededC;
            } else if (C_val != neededC) {
                cout << "No\n";
                return 0;
            }
        } else {
            // op_j = base_op[j] must equal d[i]
            if (base_op[j] != d[i]) {
                cout << "No\n";
                return 0;
            }
        }
    }

    // The final target a-value must be at least max(a[i]) so that final HP >= max(H).
    ll q = 0;
    for (int i = 1; i <= N; i++) {
        q = max(q, a[i]);
    }

    if (C_assigned) {
        // Check that the assigned C meets the lower bounds.
        if (C_val < q || C_val < C_min) {
            cout << "No\n";
        } else {
            cout << "Yes\n";
        }
    } else {
        // No exact tail constraint on C, so we can choose C = max(q, C_min)
        // which will satisfy all op_j >= 0.
        ll C_pick = max(q, C_min);
        // It's always possible in this branch.
        cout << "Yes\n";
    }

    return 0;
}
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