C

小红拿到了一个字符串，其中一定包含连续子串xiao，和连续子串hong

请你将字符串重排，使得该字符串包含xiaohong的连续子串

思路

在原串中删除xiao和hong

void solve(){
    string s;
    cin >> s;
    
    s.erase(s.find("xiao"), 4);
    s.erase(s.find("hong"), 4);
    
    cout << s + "xiaohong" << endl;
}

D

小红拿到了一个数组，她定义f(i)为删除第i个元素后，数组的中位数。现在小红想让你求出f(1), f(2), .., f(n)

思路

分类讨论

void solve(){
    int n;
    cin >> n;
    
    ve<pii> g(n + 1);
    g[0] = {0, 0};
    for(int i = 1; i <= n; i++){
        int x;
        cin >> x;
        g[i] = {x, i};
    }
    
    ve<double> res(n + 1);
    
    sort(all(g));
    
    if (n & 1) {
        double a = (1.0 * g[n / 2].x + g[n / 2 + 1].x) / 2;
        double b = (1.0 * g[n / 2 + 1].x + g[n / 2 + 2].x) / 2;
        double c = (1.0 * g[n / 2].x + g[n / 2 + 2].x) / 2;
        for (int i = 1; i <= n; i++) {
            if (i <= n / 2) res[g[i].y] = b;
            else if (i == n / 2 + 1) res[g[i].y] = c;
            else res[g[i].y] = a;
        }
    } else {
        double a = 1.0 * g[n / 2 + 1].x;
        double b = 1.0 * g[n / 2].x;
        for (int i = 1; i <= n; i++) {
            if (i <= n / 2) res[g[i].y] = a;
            else res[g[i].y] = b;
        }
    }
    
    for (int i = 1; i <= n; i++) printf("%.1lf\n", res[i]);
}

E

小红希望你构造一个数组，给你一个x，满足以下三个条件：

1. 数组的元素都是素数
2. 数组所有元素相乘恰好等于x
3. 数组任意相邻两个元素不等

思路

1. 对x质因数分解，获取质因数和指数，创建优先队列保存
2. 优先队列交替输出出现次数最大和次大的因子
3. 特判1

void solve(){
    ll x;
    cin >> x;
    
    if (x == 1) {
        cout << -1 << endl;
        return;
    }
    
    priority_queue<pii> heap;
    ve<ll> res;
    for (ll i = 2; i <= x / i; i++) {
        if (x % i == 0) {
            int t = 0;
            while (x % i == 0) {
                x /= i;
                t++;
            }
            heap.push({t, i});
        }
    }
    
    if (x > 1) heap.push({1, x});
    
    while (!heap.empty()){
        auto a = heap.top();
        heap.pop();
        a.x--;
        
        if (!heap.empty()) {
            auto b = heap.top();
            heap.pop();
            b.x--;
            res.ps(a.y), res.ps(b.y);
            if (a.x > 0) heap.push(a);
            if (b.x > 0) heap.push(b);
        } else {
            if (a.x) {
                cout << -1 << endl;
                return;
            } else res.ps(a.y);
        }
    }
    
    cout << res.size() << endl;
    for (auto x : res) cout << x << " ";
}