题解 | #Kevin的矩阵#

没做出来, 看的猫猫的题解

计算将长度为的段的时间复杂度是, 外层循环时间复杂度是, 因此总的时间复杂度是, 可以通过

枚举列的大小, 假设是, 那么产生的贡献就是

#include <bits/stdc++.h>

#define x first
#define y second
#define all(x) x.begin(), x.end()

using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using PLD = pair<LD, LD>;

const int N = 1e5 + 10, MOD = 998244353;
const int INF = 1e9;
const LL LL_INF = 1e18;
const LD EPS = 1e-8;
const int dx4[] = {-1, 0, 1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};

istream &operator>>(istream &is, i128 &val) {
    string str;
    is >> str;
    val = 0;
    bool flag = false;
    if (str[0] == '-') flag = true, str = str.substr(1);
    for (char &c : str) val = val * 10 + c - '0';
    if (flag) val = -val;
    return is;
}

ostream &operator<<(ostream &os, i128 val) {
    if (val < 0) os << "-", val = -val;
    if (val > 9) os << val / 10;
    os << static_cast<char>(val % 10 + '0');
    return os;
}

bool cmp(LD a, LD b) {
    if (fabs(a - b) < EPS) return 1;
    return 0;
}

struct Hash {
    vector<int> h, p;
    int B = 131;
    Hash(const string &s) {
        int n = s.size();
        h.resize(n + 1, 0);
        p.resize(n + 1, 1);
        for (int i = 0; i < n; ++i) {
            p[i + 1] = p[i] * B % MOD;
            h[i + 1] = (h[i] * B + s[i]) % MOD;
        }
    }

    LL get(int l, int r) {
        LL v = h[r] - h[l - 1] * p[r - l + 1] % MOD;
        v = (v % MOD + MOD) % MOD;
        return v;
    }
};

void solve() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<int> a(n);
    unordered_set<int> st;
    for (int i = 0; i < n; ++i) cin >> a[i], st.insert(a[i]);

    if (m >= n) {
        if (st.count(k)) {
            cout << 0 << '\n';
        }
        else cout << 1 << '\n';
        return;
    }

    auto calc = [&](LL c) {
        LL cur = INF;
        for (int i = 0; i < c; ++i) {
            LL t = 0;
            for (int j = i; j < n; j += c) {
                if (a[j] != k) t++;
            }
            cur = min(cur, t);
        }
        return cur;
    };

    LL ans = INF;
    for (int i = m; i <= n && i - m < ans; ++i) {
        ans = min(ans, calc(i) + i - m);
    }
    for (int i = m - 1; i > 0 && m - i < ans; --i) {
        ans = min(ans, calc(i) + m - i);
    }

    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T;
    cin >> T;
    while (T--) solve();
    cout << fixed << setprecision(15);

    return 0;
}