C. Cherry Picking

这道题用了一个类似 ODT 的题思路。

首先我们可以想到是，如果删除某些选手，只有可能会导致区间的合并，不会导致区间的分裂。所以我们可以枚举一下$x $的值，然后找到需要删除的点。用set​维护相同且相邻区间，找到删除点所在的区间后，给区间长度减一。如果区间长度为空后，就将该区间两侧的区间进行合并。同时在维护一下当前所有为 1 的区间长度。

#include <bits/stdc++.h>

using namespace std;


using ldb = long double;
using ll = long long;

using vi = vector<int>;


const int inf = 1e9;

struct Seg {
    int l, r, tag;
    mutable int len;

    Seg(int l = 0, int r = 0, int tag = 0, int len = 0)
            : l(l), r(r), tag(tag), len(len) {};

    bool operator<(const Seg &b) const {
        return l < b.l;
    }
};


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

    int n, k;
    cin >> n >> k;
    vector<array<int, 3>> a(n);// {val, idx, tag}
    for (int i = 0; i < n; i++)
        cin >> a[i][0], a[i][1] = i;

    string s;
    cin >> s;
    for (int i = 0; i < n; i++)
        a[i][2] = (s[i] == '1');


    set<Seg> seg;
    multiset<int> cnt;

    for (int i = 1, l = 0, tag = a[l][2]; i <= n; i++) {
        if (i == n) {
            seg.emplace(l, i - 1, tag, i - l);
            if (tag == 1) cnt.insert(i - l);
        } else if (a[i][2] != tag) {
            seg.emplace(l, i - 1, tag, i - l);
            if (tag == 1) cnt.insert(i - l);
            l = i, tag = a[l][2];
        }
    }

    sort(a.begin(), a.end());

    int res = 0;
    for (int x = 1, i = 0; x <= 2e5; x++) {
        if (cnt.empty() or seg.empty()) break;
        while (i < n and a[i][0] < x) {
            auto [val, idx, tag] = a[i];
            auto it = prev(seg.upper_bound(Seg(idx)));
            assert(it->l <= idx and idx <= it->r and it->tag == tag);
            if (tag == 1)
                cnt.erase(cnt.find(it->len));
            it->len--;
            if (tag == 1 and it->len > 0)
                cnt.insert(it->len);

            if (it->len == 0) {
                Seg cur(inf, -inf, -1, 0);
                if (it != seg.begin()) {
                    auto L = prev(it);
                    cur.l = min(cur.l, L->l);
                    cur.r = max(cur.r, L->r);
                    cur.len += L->len;
                    cur.tag = L->tag;

                    if (L->tag == 1) cnt.erase(cnt.find(L->len));
                    seg.erase(L);
                }
                if (next(it) != seg.end()) {
                    auto R = next(it);
                    cur.l = min(cur.l, R->l);
                    cur.r = max(cur.r, R->r);
                    cur.len += R->len;
                    cur.tag = R->tag;

                    if (R->tag == 1) cnt.erase(cnt.find(R->len));
                    seg.erase(R);
                }
                seg.erase(it);
                if (cur.tag != -1) {
                    seg.insert(cur);
                    if (cur.tag == 1) cnt.insert(cur.len);
                }
            }
            i++;
        }

        if (not cnt.empty() and *cnt.rbegin() >= k)
            res = max(res, x);
    }

    cout << res << "\n";

    return 0;
}

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签到题

#include <bits/stdc++.h>

using namespace std;

#define ll long long

using vi = vector<int>;


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

	int n;
	cin >> n;
	if(n < 10) cout << n + 1;
	else cout << 10;
	return 0;
}

K. Tasks and Bugs

简单模拟题

#include <bits/stdc++.h>

using namespace std;


using ldb = long double;
using ll = long long;

using vi = vector<int>;


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

    string s;
    map<int, vector<int>> e;

    while (getline(cin, s)) {
        s += "@";
        vector<int> a;
        int cnt = 0, f = 0;
        for (auto i: s) {
            if (isdigit(i)) {
                f = 1;
                cnt = cnt * 10 + i - '0';
            } else {
                if (f) a.push_back(cnt);
                cnt = f = 0;
            }
        }
        for (int i = 1; i < a.size(); i++)
            e[a[i]].push_back(a.front());
    }
  
    for (auto [id, v]: e) {
        cout << "CS-" << id << ": ";
        for (int f = 0; auto y: v) {
            if (f) cout << ", ";
            cout << "CS-" << y, f = 1;
        }
        cout << "\n";
    }
    return 0;
}

O. Mysterious Sequence

首先通过二维 dp，求出$X_n = a X_1 + b X_2 \(中的\)a,b\(然后解方程求出\)X_2 $，然后再重新递推出整个序列。

#include <bits/stdc++.h>

using namespace std;


using ldb = long double;
using ll = long long;
using vi = vector<int>;


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

	cout << fixed << setprecision(20);

	ldb A, B;
	cin >> A >> B;

	int n;
	cin >> n;

	vector<ldb> a(n + 1);
	cin >> a[1] >> a[n];


	if(n == 2){
		for(int i = 1; i <= n; i ++)
			cout  << a[i] << "\n";
		return 0;
	}

	vector<array<ldb,2>> k(n + 1);
	k[1] = {1, 0};
	k[2] = {0, 1};
	for(int i = 3; i <= n; i ++){
		k[i][0] = k[i - 1][0] * A + k[i - 2][0] * B;
		k[i][1] = k[i - 1][1] * A + k[i - 2][1] * B;
	}

	a[2] = (a[n] - k[n][0] * a[1]) / k[n][1];

	for(int i = 3; i < n ; i ++)
		a[i] = A * a[i - 1] + B * a[i - 2];

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