The 2024 CCPC Online Contest

https://codeforces.com/gym/105336

B - 軍訓 II

排序後肯定是最優解，方案數就是能排成有序序列的個數

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;
using i128 = __int128;

#define int i64

using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

const i64 mod = 998244353;

struct mint {
    i64 x;

    mint(i64 x = 0) : x(x) {};

    mint &operator=(i64 o) { return x = o, *this; }

    mint &operator+=(mint o) { return (x += o.x) >= mod && (x -= mod), *this; }

    mint &operator-=(mint o) { return (x -= o.x) < 0 && (x += mod), *this; }

    mint &operator*=(mint o) { return x = (x * o.x) % mod, *this; }

    mint &operator^=(int b) {
        mint w = *this, ret(1);
        for (; b; b >>= 1, w *= w) if (b & 1) ret *= w;
        return x = ret.x, *this;
    }

    mint &operator/=(mint o) { return *this *= (o ^= (mod - 2)); }

    friend mint operator+(mint a, mint b) { return a += b; }

    friend mint operator-(mint a, mint b) { return a -= b; }

    friend mint operator*(mint a, mint b) { return a *= b; }

    friend mint operator/(mint a, mint b) { return a /= b; }

    friend mint operator^(mint a, i64 b) { return a ^= b; }

    i64 val() {
        x = (x % mod + mod) % mod;
        return x;
    }
};


i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n;
    cin >> n;
    vi a(n);
    for (auto &i: a) cin >> i;
    ranges::sort(a, greater<>());
    int res = 0;
    for (int cnt = n, sum = accumulate(a.begin(), a.end(), 0); auto i: a) {
        cnt--, sum -= i;
        res += cnt * i - sum;
    }
    cout << res;
    vector<mint> fact(n + 1);
    fact[0] = 1;
    for (int i = 1; i <= n; i++)
        fact[i] = fact[i - 1] * i;
    mint ret(2 - (a.back() == a.front()));
    a.push_back(-1);
    for (int lst = -1, cnt = 0; auto i: a) {
        if (lst != i) {
            ret *= fact[cnt], cnt = 1, lst = i;
        } else {
            cnt++;
        }
    }
    cout << " " << ret.val();
    return 0;
}

D - 編碼器-解碼器

\(f[i][l][r]\)表示\(S_i\)可以匹配\(T[l,r]\)的方案數，然後按照題目的意思進行\(O(N^4)\)的轉移就好了。

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;
using i128 = __int128;

#define int i64

using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

const i64 mod = 998244353;

struct mint {
    i64 x;

    mint(i64 x = 0) : x(x) {};

    mint &operator=(i64 o) { return x = o, *this; }

    mint &operator+=(mint o) { return (x += o.x) >= mod && (x -= mod), *this; }

    mint &operator-=(mint o) { return (x -= o.x) < 0 && (x += mod), *this; }

    mint &operator*=(mint o) { return x = (x * o.x) % mod, *this; }

    mint &operator^=(int b) {
        mint w = *this, ret(1);
        for (; b; b >>= 1, w *= w) if (b & 1) ret *= w;
        return x = ret.x, *this;
    }

    mint &operator/=(mint o) { return *this *= (o ^= (mod - 2)); }

    friend mint operator+(mint a, mint b) { return a += b; }

    friend mint operator-(mint a, mint b) { return a -= b; }

    friend mint operator*(mint a, mint b) { return a *= b; }

    friend mint operator/(mint a, mint b) { return a /= b; }

    friend mint operator^(mint a, i64 b) { return a ^= b; }

    i64 val() {
        x = (x % mod + mod) % mod;
        return x;
    }
};


i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    string s, t;
    cin >> s >> t;
    int n = s.size(), m = t.size();
    vector f(n, vector(m + 1, vector(m + 1, mint())));
    for (int i = 0; i < m; i++)
        if (s[0] == t[i]) f[0][i][i] = 1;
    for (int i = 1; i < n; i++)
        for (int l = 0; l < m; l++)
            for (int r = l; r < m; r++) {
                f[i][l][r] = f[i - 1][l][r] * 2;
                if (l == r) {
                    if (s[i] == t[l]) f[i][l][r] += 1;
                } else {
                    if (s[i] == t[l]) f[i][l][r] += f[i - 1][l + 1][r];
                    if (s[i] == t[r]) f[i][l][r] += f[i - 1][l][r - 1];
                    for (int k = l; k < r; k++)
                        f[i][l][r] += f[i - 1][l][k] * f[i - 1][k + 1][r];
                    for (int k = l + 1; k < r; k++)
                        if (s[i] == t[k])
                            f[i][l][r] += f[i - 1][l][k - 1] * f[i - 1][k + 1][r];
                }
            }

    cout << f[n - 1][0][m - 1].val();

    return 0;
}

E. 隨機過程

最大點數就是使得每一層儘可能的填滿。

\[\sum \min(n, 26^i) \]

對於期望點數，同一層的點是相互獨立的。每層能夠出現的點有\(26^i\)種，對於某一種點，被選到的機率就是\(\frac{1}{26^i}\)，那麼選不到的機率就是\(1 - \frac{1}{26^i}\)，我們要選擇\(n\)次，\(n\)次都選不到的機率是\((1- \frac{1}{26^i})^ n\)，所以這個點被選中的機率就是\(1 - (1- \frac{1}{26^i})^ n\)。這層的點共有\(26^i\)個，因此總的期望就是\(26^i(1 - (1- \frac{1}{26^i})^ n)\)，共\(m\)層，所以答案就是

\[\sum 26^i[1 - (1- \frac{1}{26^i})^ n] \]

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using ui32 = uint32_t;
using i64 = long long;

//#define int i64

using vi = vector<int>;
using pii = pair<int, int>;

const i64 mod = 998244353;

struct mint {
    i64 x;

    mint(i64 x = 0) : x(x) {}

    mint &operator=(i64 o) { return x = o, *this; }

    mint &operator+=(mint o) { return (x += o.x) >= mod && (x -= mod), *this; }

    mint &operator-=(mint o) { return (x -= o.x) < 0 && (x += mod), *this; }

    mint &operator*=(mint o) { return x = (i64) x * o.x % mod, *this; }

    inline mint &operator^=(int b) {
        mint w = *this;
        mint ret(1);
        for (; b; b >>= 1, w *= w) if (b & 1) ret *= w;
        return x = ret.x, *this;
    }

    mint &operator/=(mint o) { return *this *= (o ^= (mod - 2)); }

    friend mint operator+(mint a, mint b) { return a += b; }

    friend mint operator-(mint a, mint b) { return a -= b; }

    friend mint operator*(mint a, mint b) { return a *= b; }

    friend mint operator/(mint a, mint b) { return a /= b; }

    friend mint operator^(mint a, int b) { return a ^= b; }

    i64 val() {
        x = (x % mod + mod) % mod;
        return x;
    }
};


i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;

    const mint b(26), one(1);

    mint sum;
    for (int i = 0, x = 1, t = 26; i <= m; i++) {
        sum += x;
        x *= t;
        if (x > n) x = n, t = 1;
    }

    mint res;
    for (int i = 0; i <= m; i++) {
        mint t = b ^ i;
        res += t * (one - ((one - (one / t)) ^ n));
    }

    cout << sum.val() << " " << res.val();
    return 0;
}

J. 找最小

首先我們可以求出原本的異或和\(A,B\)，再求出\(c_i = a_i \oplus b_i\)

此時我們就可以考慮用\(c_i\)選擇一些數去異或，看能否使得答案減小。

我們貪心的從高位開始操作。

如果當前位\(A,B\)都是\(1\)，能操作就操作。

如果當前位都是\(0\)，不需要任何操作。

如果當前位的較大數是\(1\)，我們可以考慮操作。

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using ui32 = uint32_t;
using i64 = long long;

//#define int i64

using vi = vector<int>;
using pii = pair<int, int>;

#define lowbit(x) ( x & -x )

struct Basis {
    vector<ui32> B;

    Basis() {
        B = vector<ui32>();
    }

    void insert(ui32 x) {
        for (auto b: B)
            x = min(x, b ^ x);
        if (x == 0) return;
        for (auto &b: B)
            b = min(b, b ^ x);
        B.push_back(x);
        return;
    }

    ui32 find(ui32 x) {
        for (auto i: B) {
            int y = i;
            while (lowbit(y) != y) y -= lowbit(y);
            if (x == y) return i;
        }
        return 0;
    }
};

void solve() {
    int n;
    cin >> n;
    vector<ui32> a(n), b(n);
    ui32 A = 0, B = 0;
    for (auto &i: a) cin >> i, A ^= i;
    for (auto &i: b) cin >> i, B ^= i;

    ui32 res = max(A, B);
    Basis basis;
    for (int i = 0; i < n; i++)
        basis.insert(a[i] ^ b[i]);

    for (ui32 i = (1 << 30); i > 0; i >>= 1) {
        if (A < B) swap(A, B);
        if (((A & i) == 0) and ((B & i) == 0)) continue;
        if (A & i) {
            ui32 x = basis.find(i);
            A ^= x, B ^= x;
        }
    }
    res = min(res, max(A, B));
    cout << res << "\n";
    return;
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int T;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

K - 取沙子游戲

這道題目是賽時先寫出SG函式，然後找規律發現。

1. \(n\)為奇數先手必勝，因為只要先手拿\(1\)後手就只能拿\(1\)
2. \(k \ge n\)先手必勝，因為先手可以一步全部拿完。
3. 剩下的情況透過找規律發現，對於\(n\)找到最大的\(t\)滿足\(t | n\) 且\(t\)為 \(2\)的整次冪，那麼\(t \ge k\) 先手必勝

原因賽時沒想出來

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;
using i128 = __int128;

#define int i64

using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

void solve() {
    int n, k;
    cin >> n >> k;
    if (n & 1) {
        cout << "Alice\n";
    } else if (k >= n) {
        cout << "Alice\n";
    } else {
        int p = 1;
        while (n % (p * 2) == 0) p *= 2;
        if (k < p) cout << "Bob\n";
        else cout << "Alice\n";
    }
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int T;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

賽後看了題解，題解總結出來一條規律就是\(\mathrm{lowbit}(n) \le k\)先手必勝。

如果\(\mathrm{lowbit}(n) \le k\)，先手可以取一個\(\mathrm {lowbit}(n)\)，然後後手一定無法一步取完，這種情況其實和 Nim 博弈很類似。道理上來說我打表找到的規律實際上是殊途同歸了

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;
using i128 = __int128;

#define int i64

using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

#define lowbit(x) ( x & -x )

void solve() {
    int n, k;
    cin >> n >> k;
    if (lowbit(n) > k) cout << "Bob\n";
    else cout << "Alice\n";
    return;
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int T;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

L - 網路預選賽

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;
using i128 = __int128;

#define int i64

using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<string> g(n);
    for (auto &i: g) cin >> i;
    int res = 0;
    for (int i = 0; i + 1 < n; i++)
        for (int j = 0; j + 1 < m; j++) {
            if (g[i][j] == 'c' and g[i][j + 1] == 'c' and g[i + 1][j] == 'p' and g[i + 1][j + 1] == 'c')
                res++;
        }
    cout << res;
    return 0;
}