update Hello2023.md

Author: coderhare

codeforces/contest/Hello2023.md

@@ -0,0 +1,102 @@
+## A
+
+容易发现只要存在`RL`字符串即可满足要求，于是只需要判断是否存在，如果不存在的话将`LR`进行反转
+
+```cpp
+void solve() {
+    int n;
+    string s;
+    cin >> n >> s;
+    // 每盏灯左边或者右边有灯
+    //LLLRLRR
+    if (n == 1 || count(all(s), 'L') == 0 || count(all(s), 'R') == 0) {
+        cout << -1 << "\n";
+        return;
+    }
+    int idx = 0;
+    for (int i = 0; i < n-1; i++) {
+        if (s[i] == 'L' && s[i+1] == 'R') {
+            idx = i+1;
+            break;
+        }
+    }
+    cout << idx << "\n";
+}
+
+```
+
+## B
+
+{**a1,a2**,a3...ak}这样的一个序列，容易发现必须满足减去a[i] + a[i+1]之后的和为0，当n=3的时候显然不满足；
+
+当n>3的时候，通过消元发现，a[i] = a[i+2*k]，进而，对剩余的元素，进行正负的划分即可，然后可以给他们安一个值，`lcm(a,b)/a`，`lcm(a,b)/b`.
+
+```cpp
+void solve() {
+    int n;
+    cin >> n;
+    if (n == 3) {
+        cout << "NO\n";
+    }
+    else if (~n&1){
+        cout << "YES\n";
+        for (int i = 1; i <= n; i+=2) {
+            cout << 1 << " " << -1 << " ";
+        }
+        cout << "\n";
+    }
+    else {
+        cout << "YES\n";
+        ll a = (n-1)/2, b = (n-2)/2;
+        cout << lcm(a,b)/a << " ";
+        for (int i = 1; i <= n-1; i+=2) {
+            cout << -lcm(a,b)/b << " " << lcm(a,b)/a << " ";
+        }
+        cout << "\n";
+    }
+}
+```
+
+## C
+
+对于m之前的所有前缀应该满足>=0，对于m之后的所有前缀也是如此。
+
+于是可以采取从m位置向前保证前缀和< 0的操作，如果不满足的话将原先存在的堆里头最大值变成负数；
+
+往后的操作同理。 实际上就是**反悔堆**的操作
+
+```cpp
+void solve() {
+    int n, m;
+    cin >> n >> m;
+    vector <ll> a(n+1);
+    for (int i = 0; i < n; i++) {
+        cin >> a[i+1];
+    }
+    ll cur = 0;
+    int ans = 0;
+    priority_queue<int> q;
+    for (int i = m; i >= 2; i--) {
+        cur += a[i];
+        q.push(a[i]);
+        if (cur > 0) {
+            auto u = q.top(); q.pop();
+            cur -= 2 * u;
+            ans++;
+        }
+    }
+    priority_queue <int,vector<int>,greater<>> q1;
+    cur = 0;
+    for (int i = m+1; i <= n; i++) {
+        cur += a[i];
+        q1.push(a[i]);
+        if (cur < 0) {
+            auto u = q1.top(); q1.pop();
+            cur -= 2 * u;
+            ans++;
+        }
+    }
+    cout << ans << "\n";
+}
+```
+