题目链接：Here

AcWing 3805. 环形数组

签到题，循环减少出现次数，如果是 cnt[x] = 1 的话加入新的数组中

const int N = 1e3 + 10;
int cnt[N];
int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int _; for (cin >> _; _--;) {
        memset(cnt, 0, sizeof(cnt));
        int n; cin >> n;
        vector<int> a(n);
        for (int &x : a) cin >> x, cnt[x] += 1;
        vector<int> b;
        for (int i = 0; i < n; ++i) {
            if (cnt[a[i]] == 1) b.push_back(a[i]);
            cnt[a[i]] -= 1;
        }
        cout << b.size() << "\n";
        for (int x : b) cout << x << " ";
        cout << "\n";
    }
}

AcWing 3804. 构造字符串

参考了一下 Y总的代码，

主要思路在于分情况考虑

• \(k > n\) 时，找到字符串中最小的字符 c，先输出 S 然后输出 \(k - n\) 个 c
• \(k \le n\)? 时，如同背包一样逆序处理，具体请看代码理解

bool cnt[30];
int n, k; string s;
char get_min() {
    for (int i = 0; i < 26; ++i) if (cnt[i]) return i + ‘a‘;
    return -1;
}
char get_nxt(int t) {
    for (int i = t + 1; i < 26; ++i) if (cnt[i]) return i + ‘a‘;
    return -1;
}
int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int _; for (cin >> _; _--;) {
        memset(cnt, 0, sizeof(cnt));
        cin >> n >> k >> s;
        for (int i = 0; i < n; ++i) cnt[s[i] - ‘a‘] = 1;

        if (k > n) {
            cout << s;
            char c = get_min();
            for (int i = n; i < k; ++i) cout << c;
            cout << "\n";
        } else {
            string ss(k, ‘a‘);
            for (int i = k - 1; i >= 0; --i) {
                char c = get_nxt(s[i] - ‘a‘);
                if (c != -1) {
                    ss[i] = c;
                    for (int j = 0; j < i; ++j) ss[j] = s[j];
                    break;
                }
                ss[i] = get_min();
            }
            cout << ss << "\n";
        }
    }
}

AcWing 3805. 环形数组

经典 线段树 板子，注意数组大小为 \(n\) 的 \(4\) 倍

const ll N = 2e5 + 10, inf = 1e18;
int n, m, w[N];
struct node {int l, r; ll dt, mn;} tr[N << 2];
void pushup(int u) { tr[u].mn = min(tr[u << 1].mn, tr[u << 1 | 1].mn); }
void pushdown(int u) {
    auto &root = tr[u], &l = tr[u << 1], &r = tr[u << 1 | 1];
    l.dt += root.dt, l.mn += root.dt;
    r.dt += root.dt, r.mn += root.dt;
    root.dt = 0; // 更新 LZ 标记
}
void build(int u, int l, int r) {
    if (l == r) tr[u] = {l, r, 0, w[l]};
    else {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }
}
void update(int u, int l, int r, int d) {
    if (tr[u].l >= l && tr[u].r <= r) tr[u].dt += d, tr[u].mn += d;
    else {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) update(u << 1, l, r, d);
        if (r > mid) update(u << 1 | 1, l, r, d);
        pushup(u);
    }
}
ll query(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) return tr[u].mn;
    else {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        ll ans = inf;
        if (l <= mid)ans = query(u << 1, l, r);
        if (r > mid) ans = min(ans, query(u << 1 | 1, l, r));
        return ans;
    }
}
int main() {
    // cin.tie(nullptr)->sync_with_stdio(false);
    cin >> n;
    for (int i = 0; i < n; ++i) cin >> w[i];
    build(1, 0, n - 1);
    cin >> m;
    while (m--) {
        int l, r, d; char c;
        scanf("%d %d%c", &l, &r, &c);
        // cin >> l >> r >> c;
        if (c == ‘\n‘) {
            if (l <= r) cout << query(1, l, r);
            else cout << min(query(1, l, n - 1), query(1, 0, r));
            cout << "\n";
        } else {
            cin >> d;
            if (l <= r)update(1, l, r, d);
            else update(1, l, n - 1, d), update(1, 0, r, d);
        }
    }
}

AcWing 第 12 场周赛