A题：小红的最小最大

思路：

大水题

code：

inline void solve() {
     int a, b, c; cin >> a >> b >> c;
     if (min(a, b) + c > max(a, b)) cout << "YES\n";
     else cout << "NO\n";
     return;
}

B题：小红的四则运算（easy） 

思路：

大水题

code：

inline void solve() {
     int ans = 0;
     int a, b, c; cin >> a >> b >> c;
     int cur = a + b + c;
     ans = max(ans, cur);
     cur = a * b + c;
     ans = max(ans, cur);
     cur = a + b * c;
     ans = max(ans, cur);
     cur = a * b * c;
     ans = max(ans, cur);
     cur = a * (b + c);
     ans = max(ans, cur);
     cur = (a + b) * c;
     ans = max(ans, cur);
     cout << ans << endl;
	 return;
}

C题：小红的四则运算（hard） 

思路：

大水题

code：

inline void solve() {
     int ans = 0;
     int a, b, c; cin >> a >> b >> c;
     int cur = a + b + c;
     ans = max(ans, cur);
     cur = a * b + c;
     ans = max(ans, cur);
     cur = a + b * c;
     ans = max(ans, cur);
     cur = a * b * c;
     ans = max(ans, cur);
     cur = a * (b + c);
     ans = max(ans, cur);
     cur = (a + b) * c;
     ans = max(ans, cur);
     cout << ans << endl;
	 return;
}

D题：小红的因式分解 

思路：

十字相乘大法。

注意正的可以表示为两个负的相乘。

code：

inline void solve() {
     ll a, b, c;
     cin >> a >> b >> c;
     if (c == 0) {
     	cout << 1 << ' ' << 0 << ' ' << a << ' ' << b << endl;
     	return;
     }
     vector<PLL> l, r;
     function<void(ll, vector<PLL>&)> add = [&](ll x, vector<PLL>& e) {
        for (ll i = min(1ll, x); i < max(0ll, x + 1); i ++ ) {
            if (x % i == 0) {
     			e.push_back({i, x / i});
     			e.push_back({-i, -x / i});
            }
        }
     };
     add(a, l), add(c, r);
     bool ok = false;
     for (auto [l1, l2] : l) {
     	for (auto [r1, r2] : r) {
     		if (l1 * r1 + l2 * r2 == b) {
     			ok = true;
     			cout << l1 << ' ' << r2 << ' ' << l2 << ' ' << r1 << endl;
     			break;
     		}
     	}
     	if (ok) break;
     }
     if (!ok) cout << "NO\n";
	 return;
}

E题：小红的树上移动 

思路：

学过期望和逆元应该就会做了。

期望等于各个值乘以它的概率的和，对于此题，我们只需统计下每层的叶节点数量和节点数量即可。

inline void solve() {
	 mod = 998244353;
     int n; cin >> n;
     vector<vector<int>> e(n + 1);
     for (int i = 1; i < n; i ++ ) {
     	int u, v; cin >> u >> v;
     	e[u].push_back(v), e[v].push_back(u);
     }
     vector<int> dep(n + 1), cnt(n + 1), tot(n + 1);
     function<void(int, int)> dfs = [&](int u, int pre) {
     	dep[u] = dep[pre] + 1;
     	tot[dep[u]] += 1;
     	if (e[u].size() == 1 && u != 1) cnt[dep[u]] += 1;
     	for (auto v : e[u]) {
     		if (v == pre) continue;
     		dfs(v, u);
     	}
     };
     dfs(1, 0);
     ll ans = 0, lv = 1;
     for (int i = 1; i <= n; i ++ ) {
     	if (cnt[i]) {
     		ll gai = lv * cnt[i] % mod * inv(tot[i]) % mod;
     		ans = (ans + (i - 1) * gai % mod) % mod;
     		lv = lv * (tot[i] - cnt[i]) % mod * inv(tot[i]) % mod;
     	}
     }
     cout << ans << endl;
	 return;
}