个人题解仅供参考。

A

签到。

C++ Code

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
  
    vector<vector<char>> a(2 * n + 1, vector<char>(2 * n + 1, '.'));
  
    for (int i = 2 * n; i > n; i--) {
        a[i][0] = '@';
        a[i][2 * n] = '@';
    }
  
    for (int i = 0; i <= 2 * n; i++) {
        a[n][i] = '@';
    }
  
    for (int i = 0; i <= n; i++) {
        a[i][i + n] = '@';
    }

    for (int i = n; i >= 0; i--) {
        a[i][n - i] = '@';
    }
  
    for (int i = 0; i <= 2 * n; i++) {
        for (int j = 0; j <= 2 * n; j++) {
            cout << a[i][j];
        }
        cout << '\n';
    }
  
    return 0;
}

C

一个数要么分给惠惠，要么分给悠悠，直接 dfs 就行。

C++ Code

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

void solve() {
    int n;
    cin >> n;

    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }

    if (n % 2 == 1) {
        cout << "Yunyun\n";
        return; 
    }

    vector<int> p, q;
    bool ok = 0;
    function<void(int)> dfs = [&](int I) {
        if (ok) {
            return;
        }
        int len1 = p.size(), len2 = q.size();
        for (int i = 0; i < min(len1, len2); i++) {
            if (p[i] != q[i]) {
                return;
            }
        }
        if (len1 == n / 2 && len2 == n / 2) {
            ok = 1;
            return;
        }
        if (I == n) {
            return;
        }
        if (len1 > n / 2) {
            return;
        }
        if (len2 > n / 2) {
            return;
        }        
        p.push_back(a[I]);
        dfs(I + 1);
        p.pop_back();
        q.push_back(a[I]);
        dfs(I + 1);
        q.pop_back();
    };
  
    p.push_back(a[0]);
    dfs(1);
  
    cout << (ok ? "Megumin" : "Yunyun") << '\n';
  
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;
  
    while (t--) {
        solve();
    }

    return 0;
}

E

这里使用字符串哈希。显然此题还有其他别的方法。

前后一样的消掉，然后枚举翻转点，两种情况，翻转前面或者翻转后面。数据 $5\cdot 10^{5}$ ，单哈希可能会哈希碰撞。时间复杂度 $\mathcal{O}(n)$ 。

与此题类似：https://atcoder.jp/contests/abc284/tasks/abc284_f

赛后发现此题来自：https://codeforces.com/gym/104095/problem/K

C++ Code

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

constexpr int B = 777;
constexpr i64 P = 100000000000031;

i64 *p;

void init(int N) {
    p = new i64 [N + 1];
    p[0] = 1;
    for (int i = 1; i <= N; i++) {
        p[i] = p[i - 1] * B % P;
    }
}

struct StringHash {
    vector<i64> h;
    StringHash() : h(1) {}
    void push_back(char ch) {
        h.push_back((h.back() * B + ch) % P);
    }
    i64 get(int l, int r) {
        return (h[r] + __int128(h[l]) * (P - p[r - l])) % P;
    }
};

void solve() {
    string s;
    cin >> s;
    int n = s.size();
    deque<char> q;
    for (int i = 0; i < n; i++) {
        q.push_back(s[i]);
    }
    while (q.size() > 1 && q.front() == q.back()) {
        q.pop_back();
        q.pop_front();
    }
    if (q.size() < 2) {
        cout << "Yes\n";
        return;
    }
    StringHash hs, hs1;
    int m = q.size();
    for (int i = 0; i < m; i++) {
        hs.push_back(q[i]);
    }
    for (int i = m - 1; i >= 0; i--) {
        hs1.push_back(q[i]);
    }
    auto work = [&](StringHash &hs, StringHash &hs1) {
        for (int len = 1; len <= m; len++) {
            int i = 0, j = i + len;
            if (!(j + len <= m && j <= m)) {
                continue;
            }
            if (hs.get(i, j) == hs.get(j, j + len) && 
                hs.get(j + len, m) == hs1.get(0, m - j - len)) {
                return 1;
            }
            int i0 = m - len, j0 = m;
            if (j <= i0 && hs.get(i, j) == hs.get(i0, j0) && 
                hs.get(j, i0) == hs1.get(m - i0, m - j)) {
                return 1;
            }
        }
        return 0;
    };

    if (work(hs, hs1) || work(hs1, hs)) {
        cout << "Yes\n";
        return;
    }
    cout << "No\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    init(5E5);

    int t;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

F

二维前缀和，朴素解法 $\mathcal{O}(n^{4})$ ，用双指针可以优化为 $\mathcal{O}(n^{3})$ ，这里用的是枚举矩形宽，二分矩形长，复杂度 $\mathcal{O}(n^{3}\log n)$ 。

C++ Code

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m, k;
    cin >> n >> m >> k;

    vector<vector<int>> a(n + 1, vector<int>(m + 1));

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            char ch;
            cin >> ch;
            if (ch == '.') {
                a[i][j] = 1;
            }
        }
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            a[i][j] += a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];
        }
    }

    auto get = [&](int x1, int y1, int x2, int y2) {
        return a[x2][y2] + a[x1][y1] - a[x2][y1] - a[x1][y2];
    };

    int ans = 1E9;
    
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= m; j++) {
            for (int W = 0; i + W <= n; W++) {
                int l = 0, r = m - j;
                auto check = [&](int H) {
                    return get(i, j, i + W, j + H) >= k;
                };
                int H = -1;
                while (l <= r) {
                    int mid = (l + r) >> 1;
                    if (check(mid)) {
                        r = mid - 1;
                        H = mid;
                    } else {
                        l = mid + 1;
                    }   
                }
                if (H != -1) {
                    ans = min(ans, W * H);
                }
            }
        }
    }

    if (ans == 1E9) {
        cout << "No Solution\n";
    } else {
        cout << ans << '\n';
    }

    return 0;
}

G

最大生成树。

C++ Code

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

struct UnionFind {
    int n;
    vector<int> f;
    UnionFind(const int &n) : n(n), f(n) {
        iota(f.begin(), f.end(), 0);
    }
    int get(int x) {
        return x == f[x] ? x : f[x] = get(f[x]);
    }
    bool unite(int x, int y) {
        int gx = get(x), gy = get(y);
        if (gx != gy) {
            f[gx] = gy;
            return 1;
        }
        return 0;
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m;
    cin >> n >> m;
    UnionFind f(n);
    i64 ans = 0;
    vector<array<int, 3>> g;
    for (int i = 0; i < m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        u--;
        v--;
        g.push_back({w, u, v});
        ans += w;
    }

    sort(g.begin(), g.end(), greater());
    for (int i = 0; i < m; i++) {
        if (f.unite(g[i][1], g[i][2])) {
            ans -= g[i][0];
        }
    }
    cout << ans << '\n';

    return 0;
}

I

一直对折，对折到差不多了就再随便折一次。

C++ Code

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

void solve() {
    int w, h, x, y;
    cin >> w >> h >> x >> y;

    auto work = [&](int now, int aim) {
        if (now < aim) {
            return (int) 1E9;
        }
        if (now == aim) {
            return 0;
        }
        int cnt = 0;
        int tmp = now;
        while ((tmp + 1) / 2 > aim) {
            cnt++;
            tmp = (tmp + 1) / 2;
        }
        return cnt + 1;
    };
    
    int ans = min(work(w, x) + work(h, y), work(w, y) + work(h, x));
    if (ans >= 1E9) {
        cout << -1 << '\n';
        return;
    }
    cout << ans << '\n';

}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;
  
    while (t--) {
        solve();
    }

    return 0;
}

J

$n= k_1x m$

$y=k_2n m$

$y=k_1k_2x k_2m m$

$k_2m m=y\mod x$

令 $k_2=1$ ，则 $m=\frac{y\mod x}{2}$

$n=y-m$

C++ Code

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

void solve() {
    int x, y;
    cin >> x >> y;

    if (x > y) {
        cout << x + y << '\n';
        return;
    }
    cout << y - (y % x) / 2 << '\n';

}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;
  
    while (t--) {
        solve();
    }

    return 0;
}

L

签到。

题解 | #西安理工大学2023年程序设计竞赛#

A

C

E

F

G

I

J

L