题解 | #武汉工程大学第六届ACM程序设计竞赛（同步赛）#

C

dfs。

$O(n)$ 。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

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

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

    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);
    }

    vector<i64> sum(n), d(n), ans(n);
    function<void(int, int, int)> dfs = [&](int x, int p, int dep) {
        sum[x] = a[x];
        for (auto y : g[x]) {
            if (y != p) {
                dfs(y, x, dep + 1);
                sum[x] += sum[y];
            }
        }
        if (dep - k > 0) {
            d[dep - k - 1] += sum[x];
        }
        ans[x] += sum[x] - d[dep];
        d[dep] = 0;
    };

    dfs(0, -1, 0);

    for (int i = 0; i < n; i++) {
        cout << ans[i] << " \n"[i == n - 1];
    }

    return 0;
}

D

单调栈，前缀和。

$O(n)$ 。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

constexpr int P = 998244353;

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 work = [&]() {
        vector<i64> sum0(n + 1), sum1(n + 2);
        for (int i = 0; i < n; i++) {
            sum0[i + 1] = ((sum0[i] + a[i]) % P + P) % P;
        }
        for (int i = 0; i <= n; i++) {
            sum1[i + 1] = ((sum1[i] + sum0[i]) % P + P) % P;
        }

        vector<int> l(n), r(n), st;
        for (int i = 0; i < n; i++) {
            while (!st.empty() && a[st.back()] < a[i]) {
                st.pop_back();
            }
            l[i] = st.empty() ? -1 : st.back();
            st.push_back(i);
        }

        st = {};

        for (int i = n - 1; i >= 0; i--) {
            while (!st.empty() && a[st.back()] <= a[i]) {
                st.pop_back();
            }
            r[i] = st.empty() ? n : st.back();
            st.push_back(i);     
        }

        i64 res = 0;
        for (int i = 0; i < n; i++) {
            i64 v = (a[i] % P + P) % P;
            i64 sum = ((sum1[r[i] + 1] - sum1[i + 1] + P) % P * (i - l[i]) % P - (sum1[i + 1] - sum1[l[i] + 1] + P) % P * (r[i] - i) % P + P) % P;

            res = (res + v * sum % P) % P;
        }
        return res;
    };

    i64 ans0 = work();

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

    i64 ans1 = work();

    cout << (ans0 - ans1 + P) % P << '\n';

    return 0;
}

E

对每个 $i$ 合法区间 $[l_i,r_i)$ 加起来。

$O(n\log n)$ 。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

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

    int n, l, r;
    cin >> n >> l >> r;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    sort(a.begin(), a.end());
    
    i64 ans = 0;
    for (int i = 0; i < n; i++) {
        int li = lower_bound(a.begin(), a.end(), l - a[i]) - a.begin();
        int ri = upper_bound(a.begin(), a.end(), r - a[i]) - a.begin();
        ans += ri - li - (li <= i && i < ri);
    }
    cout << ans / 2 << '\n';

    return 0;
}

F

$O(n^{2})$ 。

#include "bits/stdc++.h"

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    vector<int> sum(n + 1);
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        sum[i + 1] = sum[i] + (s == "byl");
    }

    int ans = 0;
    for (int l = 0; l < n; l++) {
        for (int r = l + 1; r <= n; r++) {
            ans += (r - l - sum[r] + sum[l] < sum[r] - sum[l]);
        }
    }
    cout << ans << '\n';

    return 0;
}

G

使用 $\text{__gnu_pbds::tree}$ 。

$O(n\log n)$ 。

#include "bits/stdc++.h"

using namespace std;
using i64 = int64_t;

#include "ext/pb_ds/assoc_container.hpp"

using namespace __gnu_pbds;

template <typename Key, typename Cmp_Fn = less<Key>>
class Set : public tree<Key, null_type, Cmp_Fn, rb_tree_tag,
                        tree_order_statistics_node_update> {};

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

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

    int x = 0, y = 0;
    Set<pair<i64, int>> s;
    s.insert({0, 0});
    i64 ans = 0;
    for (int i = 0; i < n; i++) {
        string t;
        cin >> t;
        t == "byl" ? x++ : y++;

        ans += s.order_of_key({x - i64(k) * y, -1});
        s.insert({x - i64(k) * y, i + 1});
    }
    cout << ans << '\n';

    return 0;
}

H

构造的数组只含有 $1,2,3$ 。

$O(n)$ 。

#include "bits/stdc++.h"

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n, k, s;
    cin >> n >> k >> s;
    
    if (n > 1 && k == 1) {
        cout << "NO\n";
        return 0;
    }
    
    vector<int> a(n);
    for (int i = 0; i < n; i += k) {
        for (int j = 0; j < k && i + j < n; j++) {
            a[i + j] = (j % 2) + 1;
        }
        if (k % 2 == 1 && i > 1) {
            a[i - 1] = 3;
        }
    }

    if (accumulate(a.begin(), a.end(), 0LL) > s) {
        cout << "NO\n";
        return 0;
    }

    cout << "YES\n";
    for (int i = 0; i < n; i++) {
        cout << a[i] << " \n"[i == n - 1];
    }

    return 0;
}

J

全排列。

#include "bits/stdc++.h"

using namespace std;

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

    vector a(7, vector<int>(6));
    for (int i = 0; i < 7; i++) {
        for (int j = 0; j < 6; j++) {
            cin >> a[i][j];
            a[i][j]--;
        }
    }

    vector<int> p(7);
    iota(p.begin(), p.end(), 0);

    do {
        vector<int> b;
        for (int i = 0; i < 6; i++) {
            for (int j = 0; j < 6; j++) {
                if (a[p[i]][j] == a[p[6]][i]) {
                    b.push_back(a[p[i]][(j + 1) % 6]);   
                    b.push_back(a[p[i]][(j - 1 + 6) % 6]);        
                    break;
                }
            }
        }
        int cnt = 0;
        for (int i = 0; i < 12; i++) {
            cnt += (b[i] == b[(i + 1) % 12]);
        }
        if (cnt == 6) {
            cout << "YES\n";
            return 0;
        }
    } while (next_permutation(p.begin(), p.end()));

    cout << "NO\n";

    return 0;
}