牛客周赛62

A - E

A：

void solve() {
    string s;cin >> s;
    swap(s[0], s[1]);
    cout << s << '\n';
}

B：

#define int long long
void solve() {
    int n, x;cin >> n >> x;
    vector<int> a(n + 1);
    for (int i = 1;i <= n;i++)cin >> a[i];
    int ans = 0;
    for (int i = 1;i <= n;i++) {
        if (!x)continue;
        ans += a[i];
        if (x > 0) {
            x -= a[i];
        } else if (x < 0) {
            x += a[i];
        }
    }
    cout << ans << '\n';
}

C：

看错题目力

#define int long long
constexpr double pi = 3.14159265358979324;
void solve() {
    cout << fixed << setprecision(15);
    int n, k;cin >> n >> k;
    vector<pair<double, double>> cnt(n);
    for (int i = 0;i < n;i++) {
        double x, y, r;cin >> x >> y >> r;
        double dis = x * x + y * y;
        cnt[i] = {max<double>(0,r - sqrt(dis)),pi * r * r};
    }
    sort(cnt.begin(), cnt.end(), [](auto x, auto y) {
        return x.first * x.second < y.first * y.second;
        });
    double ans = 0;
    for (int i = 0;i < n - k;i++) {
        ans += cnt[i].first * cnt[i].second;
    }
    cout << ans << '\n';
}

D：

取模需要注意

#define int long long
constexpr int mod = 1e9 + 7;
void solve() {
    int n;cin >> n;
    vector<vector<int>> g(n + 1);
    for (int i = 1;i < n;i++) {
        int u, v;cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    if (n == 1) {
        cout << "1\n";return;
    }

    vector<int> p(n + 1), q(n + 1);
    auto dfs = [&](auto self, int u, int fa, int d, int mul)->void {
        if (g[u].size() == 1 && u != 1) {
            p[u] = d;q[u] = mul;
        }
        int cnt = 0;
        for (auto v : g[u]) {
            if (v == fa)continue;
            cnt++;
        }
        for (auto v : g[u]) {
            if (v == fa)continue;
            self(self, v, u, d + 1, mul * cnt % mod);
        }
        };
    dfs(dfs, 1, 0, 1, 1);
    int ans = 0;
    for (int i = 1;i <= n;i++) {
        if (p[i] && q[i]) {
            ans += p[i] * inv(q[i]) % mod;
            ans %= mod;
        }
    }
    cout << ans << '\n';
}

E：

pbds 的第一次运用

template<typename T>
using ordered_multiset = tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>;
void solve() {
    int n, q;cin >> n >> q;
    vector<int> a(n + 1);
    for (int i = 1;i <= n;i++) cin >> a[i];
    q--;int l, r;cin >> l >> r;
    int k = r - l + 1;

    ordered_multiset<int> s;
    vector<int> ans;
    ans.push_back(0);

    for (int i = 1;i <= k;i++) {
        s.insert(a[i]);
    }
    ans.push_back(*s.find_by_order(k / 2));
    for (int i = k + 1;i <= n;i++) {
        s.erase(s.upper_bound(a[i - k]));
        s.insert(a[i]);
        ans.push_back(*s.find_by_order(k / 2));
    }
    cout << ans[l] << '\n';
    while (q--) {
        cin >> l >> r;
        cout << ans[l] << '\n';
    }
}

F :

主席树板子，可以理解我区间求第 $k$ 大的板子。

对于区间求第 $k$ 大相关的解决方法特别多

• 整体二分
• 主席树/树状数组
• 分块
• 莫队

P3834 【模板】可持久化线段树 2 - 洛谷 | 计算机科学教育新生态

->一个区间求第 k 小的板子(可以通过本题)

constexpr int N = 200010;
int n, q, m, cnt = 0;
int a[N], b[N], T[N];
int sum[N << 5], L[N << 5], R[N << 5];

int build(int l, int r) {
    int rt = ++cnt;
    sum[rt] = 0;
    int mid = l + r >> 1;
    if (l < r) {
        L[rt] = build(l, mid);
        R[rt] = build(mid + 1, r);
    }
    return rt;
}

int update(int pre, int l, int r, int x) {
    int rt = ++cnt;
    L[rt] = L[pre]; R[rt] = R[pre]; sum[rt] = sum[pre] + 1;
    int mid = l + r >> 1;
    if (l < r) {
        if (x <= mid) L[rt] = update(L[pre], l, mid, x);
        else R[rt] = update(R[pre], mid + 1, r, x);
    }
    return rt;
}

int query(int u, int v, int l, int r, int k) {
    if (l >= r) return l;
    int x = sum[L[v]] - sum[L[u]];
    int mid = l + r >> 1;
    if (x >= k) return query(L[u], L[v], l, mid, k);
    else return query(R[u], R[v], mid + 1, r, k - x);
}

int main() {
    int n, q;cin >> n >> q;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        b[i] = a[i];
    }
    sort(b + 1, b + 1 + n);
    m = unique(b + 1, b + 1 + n) - b - 1;
    T[0] = build(1, m);
    for (int i = 1; i <= n; i++) {
        int t = lower_bound(b + 1, b + 1 + m, a[i]) - b;
        T[i] = update(T[i - 1], 1, m, t);
    }
    while (q--) {
        int l, r;cin >> l >> r;
        int k = r - l + 1;k = (k + 1) / 2;
        int t = query(T[l - 1], T[r], 1, m, k);
        cout << b[t] << '\n';
    }
}