string solve() {cin >> n;for (int i = 0; i < n; i ++) cin >> a[i];int A = 0, B = 0, mode = 0;for (int i = n - 1; i >= 0; i --, mode ^= 1)if (mode & 1) B += a[i];else A += a[i];return A > B ? "Alice" : "Bob";
}

string solve() {cin >> n;for (int i = 0; i < n; i ++) cin >> a[i], b[i] = a[i];sort(a, a + n);return a[0] == b[0] || a[n - 1] == b[n - 1] ? yes : no;
}

雙指針

void solve() {cin >> n >> m >> s;int l = s.find('I') - 1, r = l + 2;while (m --) {string op; cin >> op;if (op[0] == 'b') {if (l != -1) {if (r != n && s[l] == '(' && s[r] == ')') r ++;l --;}} else {if (r != n) r ++;}}for (int i = 0; i <= l; i ++) cout << s[i];cout << 'I';for (int i = r; i < n; i ++) cout << s[i];cout << endl;
}

用兩個棧模擬

void solve() {cin >> n >> m >> s;int idx = s.find('I');stack<char> l, r, t;for (int i = 0; i < idx; i ++) l.push(s[i]);for (int i = n - 1; i > idx; i --) r.push(s[i]);while (m --) {string op; cin >> op;if (op[0] == 'b') {if (!l.empty()) {if (!r.empty() && l.top() == '(' && r.top() == ')') r.pop();l.pop();}} else if (op[0] == 'd') {if (!r.empty()) r.pop();} else if (op[0] == '<') {if (!l.empty()) {r.push(l.top());l.pop();}} else {if (!r.empty()) {l.push(r.top());r.pop();}}}while (!l.empty()) t.push(l.top()), l.pop();while (!t.empty()) cout << t.top(), t.pop();cout << 'I';while (!r.empty()) cout << r.top(), r.pop();cout << endl;
}

貪心，讓 $a[i]+b[i]=ma{x}_{j=1}^{i?1}{a}_j$

void solve() {cin >> n;for (int i = 0; i < n; i ++) cin >> a[i];int Max = -1e9;for (int i = 0; i < n; i ++) {b[i] = max(0, Max - a[i]);Max = max(Max, a[i]);}for (int i = 0; i < n; i ++) cout << b[i] << ' ';cout << endl;
}

由于一個數如果給 $|$ 操作只會多不會少，但是給 $\&$ 操作只會少不會多，因此貪心地讓 $\&$ 操作只包含 $a[n]$，接下來遍歷一下前兩個操作的分界點即可

ll solve() {cin >> n;for (int i = 1; i <= n; i ++) cin >> a[i], b[i] = a[i] ^ b[i - 1];ll res = 0, cnt = a[n - 1];for (int i = n - 2; i; i --) {res = max(res, cnt + b[i]);cnt |= a[i];}return res + a[n];
}

樹狀數組維護逆序對，$f[i]=nu{m}_{i|arr[k]}?f[2i]?f[3i]?\dots f[?\frac{n}{i}?i]$

其中逆序操作保證 $f[2i],f[3i],\dots \dots f[?\frac{n}{i}?i]$ 是已經求得的

ll f[N];struct BIT {int tr[N];inline int lowbit(int x) { return x & -x; }void modify(int x, int k) {while (x) tr[x] += k, x -= lowbit(x);}int query(int x) {int res = 0;while (x <= n) res += tr[x], x += lowbit(x);return res;}
} bit;ll solve() {cin >> n;for (int i = 1; i <= n; i ++) {int x; cin >> x;a[x] = i;}for (int i = n; i; i --) {for (int j = i; j <= n; j += i) {f[i] += bit.query(a[j]);bit.modify(a[j], 1);}for (int j = i; j <= n; j += i) {bit.modify(a[j], -1);if (j != i) f[i] -= f[j];}}return f[1];
}