CodeTON Round 3 (Div. 1 + Div. 2, Rated, Prizes!)1750

https://codeforces.com/contest/1750/
新手第二把solved2 rating+254

A

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
using namespace std;

void solve() {
    int n;
    cin >> n;
    vector<int> a(n + 1);
    // bool flag = true;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    if (a[1] == 1)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return;
}
signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int _ = 1;
    cin >> _;
    while (_--) {
        solve();
    }
    return 0;
}

B

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
using namespace std;

void solve() {
    int a;
    int n;
    cin >> n;
    string s;
    cin >> s;
    int ans0 = 0, ans1 = 0;
    int cnt0 = 0, cnt1 = 0;
    int s0 = 0, s1 = 0;
    // 记录连续0和1的个数
    for (auto a : s) {
        if (a == '0') {
            ans0++;
            cnt0++;
            cnt1 = 0;
        } else {
            ans1++;
            cnt1++;
            cnt0 = 0;
        }
        s0 = max(s0, cnt0);
        s1 = max(s1, cnt1);
    }
    // 0 和 1 都有
    int r1 = ans0 * ans1;
    // 只有0
    int r2 = s0 * s0;
    // 只有1
    int r3 = s1 * s1;
    cout << max({r1, r2, r3}) << endl;
    return;
}

signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int _ = 1;
    cin >> _;
    while (_--) {
        solve();
    }
    return 0;
}

C

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
#define x first
#define y second
using namespace std;

typedef pair<int, int> PII;

void solve() {
    int n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    s1 = "?" + s1;
    s2 = "?" + s2;
    // 记录不相同的个数
    int cnt = 0;
    // 记录0的个数, 1的个数
    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; i++) {
        if (s1[i] == s2[i]) {
            if (s1[i] == '0')
                cnt1++;
            else
                cnt2++;
        } else
            cnt++;
    }
    // 如果不是全相等或者全相反，就一定不可以
    if (cnt != 0 && cnt != n) {
        cout << "NO" << endl;
        return;
    }
    // 如果全是0，就不用判断了，一定可以
    if (cnt1 == n) {
        cout << "YES" << endl;
        cout << 0 << endl;
        return;
    }
    // 如果全是1，就不用判断了，一定可以
    if (cnt2 == n) {
        cout << "YES" << endl;
        cout << 2 << endl;
        cout << 1 << " " << 1 << endl;
        cout << 2 << " " << n << endl;
        return;
    }

    // 一般情况
    vector<PII> ans;
    // 记录目前操作了多少次
    int cur = 0;

    // 如果是 Case1，就不管
    // 如果是 Case2，就要多操作一下转换为 Case1
    if (cnt) cur++;

    // 对于Case1 操作每一个 1
    for (int i = 1; i <= n; i++) {
        if (s1[i] == '1') {
            ans.push_back({i, i});
            cur++;
        }
    }
    // 如果是1的个数是单数的，最后会剩下一个11对，也进行一下操作
    if (cur % 2 == 1) {
        ans.push_back({1, 1});
        ans.push_back({2, n});
        ans.push_back({1, n});
    }
    cout << "YES" << endl;
    cout << ans.size() << endl;
    for (auto it : ans) cout << it.first << " " << it.second << endl;
}
signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int _ = 1;
    cin >> _;
    while (_--) {
        solve();
    }
    return 0;
}