题解 | #2024年中国传媒大学程序设计大赛（同步赛）#

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A

没动脑。

线段树维护最大子段和。

。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

template<class Info,
        class Merge = plus<Info>>
struct SegmentTree {
    const int n;
    const Merge merge;
    vector<Info> info;
    SegmentTree(int n) : n(n), merge(Merge()), info(4 << __lg(n)) {}
    SegmentTree(vector<Info> init) : SegmentTree(init.size()) {
        function<void(int, int, int)> build = [&](int p, int l, int r) {
            if (r - l == 1) {
                info[p] = init[l];
                return;
            }
            int m = (l + r) / 2;
            build(2 * p, l, m);
            build(2 * p + 1, m, r);
            pull(p);
        };
        build(1, 0, n);
    }
    void pull(int p) {
        info[p] = merge(info[2 * p], info[2 * p + 1]);
    }
    void modify(int p, int l, int r, int x, const Info &v) {
        if (r - l == 1) {
            info[p] = v;
            return;
        }
        int m = (l + r) / 2;
        if (x < m) {
            modify(2 * p, l, m, x, v);
        } else {
            modify(2 * p + 1, m, r, x, v);
        }
        pull(p);
    }
    void modify(int p, const Info &v) {
        modify(1, 0, n, p, v);
    }
    Info rangeQuery(int p, int l, int r, int x, int y) {
        if (l >= y || r <= x) {
            return Info();
        }
        if (l >= x && r <= y) {
            return info[p];
        }
        int m = (l + r) / 2;
        return merge(rangeQuery(2 * p, l, m, x, y), rangeQuery(2 * p + 1, m, r, x, y));
    }
    Info rangeQuery(int l, int r) {
        return rangeQuery(1, 0, n, l, r);
    }
};

struct Info {
    i64 sum;
    i64 lmx;
    i64 rmx;
    i64 mx;
    Info(i64 x = 0) : sum(x), lmx(x), rmx(x), mx(x) {}
};

Info operator+(const Info &a, const Info &b) {
    Info res;
    res.sum = a.sum + b.sum;
    res.lmx = max(a.sum + b.lmx, a.lmx);
    res.rmx = max(b.sum + a.rmx, b.rmx);
    res.mx = max({a.mx, b.mx, a.rmx + b.lmx});
    return res;
}

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

    int n;
    cin >> n;

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

    SegmentTree<Info> seg(n);
    for (int i = 0; i < n; i++) {
        seg.modify(i, a[i]);
    }

    for (int i = 0; i < n; i++) {
        cout << a[i] + seg.rangeQuery(0, i).rmx + seg.rangeQuery(i + 1, n).lmx << " \n"[i == n - 1];
    }

    return 0;
}

B

打表发现一个东西，然后暴力。

。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

constexpr int N = 2E5;
int cnt[N + 1];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        cnt[a[i]]++;
    }

    i64 ans = 0;
    for (int i = 1; i <= N; i++) {
        for (int j = i; j <= N; j += i) {
            int sum = 0;
            for (int k = j; k <= N; k += j) {
                sum += cnt[k];
            }
            ans += i64(cnt[i]) * cnt[j] * sum;
        }
    }
    cout << ans << '\n';

    return 0;
}

C

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

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

    int n;
    cin >> n;

    if (n < 3) {
        cout << "-1\n";
    } else {
        cout << "CUC" << string(n - 3, 'A') << '\n';
    }

    return 0;
}

D

来自🐔哥。

我是🐔哥粉丝。

。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

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

    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
    }
    
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cout << (a[i] + b[j] > n);
        }
        cout << '\n';
    }
    
    return 0;
}

E

签到。

。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

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

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

    if ((n - 1) * (m - 1) < k) {
        cout << "-1\n";
        return 0;
    }
  
    vector a(n, vector<int>(m));
    for (int i = 0; i + 1 < n; i++) {
        for (int j = 0; j + 1 < m; j++) {
            if (k > 0) {
                for (int i0 = 0; i0 < 2; i0++) {
                    for (int j0 = 0; j0 < 2; j0++) {
                        a[i + i0][j + j0] = 1;
                    }
                }
                k--;
            }
        }
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cout << a[i][j];
        }
        cout << '\n';
    }

    return 0;
}

F

dfs 搭配分解法和剪枝。

复杂度不详。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

vector<int> minp, primes;
 
void sieve(int n) {
    minp.assign(n + 1, 0);
    
    for (int i = 2; i <= n; i++) {
        if (minp[i] == 0) {
            minp[i] = i;
            primes.push_back(i);
        }
        
        for (auto p : primes) {
            if (i * p > n) {
                break;
            }
            minp[i * p] = p;
            if (p == minp[i]) {
                break;
            }
        }
    }
}

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

    int n;
    i64 k;
    cin >> n >> k;  
    
    sieve(n);

    vector<map<int, int>> mp(n);
    vector<int> ok(n);
  
    vector<vector<int>> g(n);
    for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        u--, v--;

        g[u].push_back(v);
        g[v].push_back(u);
    }

    int ans = 0;
    function<void(int, int)> dfs = [&](int x, int p) {
        for (auto y : g[x]) {
            if (y != p) {
                dfs(y, x);
                if (ok[y]) {
                    ok[x] = 1;
                    continue;
                }
                for (auto [p, c] : mp[y]) {
                    mp[x][p] += c;
                }
            }
        }

        if (!ok[x]) {
            int x0 = x + 1;
            while (x0 > 1) {
                int p0 = minp[x0];
                x0 /= p0;
                mp[x][p0]++;
            }

            i64 res = 1;
            for (auto [p, c] : mp[x]) {
                if (res * (c + 1) >= k) {
                    ans++;
                    ok[x] = 1;
                    break;
                }
                res *= (c + 1);
            }            
        } else {
            ans++;
        }
    };

    dfs(0, -1);

    cout << ans << '\n';

    return 0;
}

H

离散化后类似双指针。

。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

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

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

    auto v = a;
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());

    for (int i = 0; i < n; i++) {
        a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
    }

    vector<int> c0(n), c1(n);
    int tot0 = 0, tot1 = 0;
    int l0 = 0, l1 = 0;
    i64 ans = 0;
    for (int r = 0; r < n; r++) {
        if (!c0[a[r]]) {
            tot0++;
        }
        c0[a[r]]++;
        if (!c1[a[r]]) {
            tot1++;
        }
        c1[a[r]]++;
        while (tot0 > 3) {
            c0[a[l0]]--;
            if (!c0[a[l0]]) {
                tot0--;
            }
            l0++;
        }
        while (tot1 > 2) {
            c1[a[l1]]--;
            if (!c1[a[l1]]) {
                tot1--;
            }
            l1++;
        }
        ans += l1 - l0;
    }
    cout << ans << '\n';

    return 0;
}

H

类似去年天梯赛 L2-4。

可以 bfs。

。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

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

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

    vector t(n + 2, vector<int>(m + 2));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            t[i + 1][j + 1] = (a[i][j] == '*');
        }
    }

    queue<pair<int, int>> q;
    q.emplace(0, 0);
    t[0][0] = 2;
    while (!q.empty()) {
        auto [x0, y0] = q.front();
        q.pop();

        for (auto [dx, dy] : vector<pair<int, int>>{{+1, +0}, {+0, +1}, {-1, +0}, {+0, -1}}) {
            int x1 = x0 + dx, y1 = y0 + dy;

            if (x1 >= 0 && x1 < n + 2 && y1 >= 0 && y1 < m + 2) {
                if (t[x1][y1] == 1) {
                    t[x1][y1] = 3;
                } else if (t[x1][y1] == 0) {
                    t[x1][y1] = 2;
                    q.emplace(x1, y1);
                }
            }
        }
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cout << (t[i + 1][j + 1] == 1 && a[i][j] == '*' ? '*' : '.');
        }
        cout << '\n';
    }

    return 0;
}

J

数学题。

考虑枚举 。

感觉写的有点假，但是过了。

。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

vector<int> minp, primes;
 
void sieve(int n) {
    minp.assign(n + 1, 0);
    
    for (int i = 2; i <= n; i++) {
        if (minp[i] == 0) {
            minp[i] = i;
            primes.push_back(i);
        }
        
        for (auto p : primes) {
            if (i * p > n) {
                break;
            }
            minp[i * p] = p;
            if (p == minp[i]) {
                break;
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    sieve(1E6);
  
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    sort(a.begin(), a.end());
  
    int mx = a[n - 1];
  
    int g = 0;
    for (int i = 1; i < n; i++) {
        g = gcd(g, a[i] - a[0]);
    }
  
    i64 ans = 0;
    for (int i = mx; i <= m; i++) {
        int g0 = g;
        g0 = gcd(i - a[0], g);

        map<int, int> mp;
        while (g0 > 1) {
            int p = minp[g0];
            g0 /= p;
            mp[p]++;
        }
        i64 t = 1;
        for (auto [u, v] : mp) {
            t *= (v + 1);
        }
        ans += t;
    }
    cout << ans << '\n';

    return 0;
}