HHKB Programming Contest 2023(AtCoder Beginner Contest 327)

HHKB Programming Contest 2023(AtCoder Beginner Contest 327)

A - ab

int main() {
    IOS;
    string s;
    cin >> n >> s;
    bool f = false;
    for (int i = 1; i < n; ++i)
        if (s[i - 1] == 'a' && s[i] == 'b') f = true;
        else if (s[i] == 'a' && s[i - 1] == 'b') f = true;
    cout << (f ? "Yes" : "No");
    return 0;
}

B - A^A

int main() {
    IOS;
    cin >> n;
    int f = -1;
    for (int i = 1, j = 1; true; ++i, j = 1) {
        ll s = 1;
        for (; j <= i && s <= n / i; ++j) s *= i;
        if (j <= i || s > n) break;
        if (s == n) { f = i; break; } 
    }
    cout << f;
    return 0;
}

C - Number Place

int a[9][9], f[9];

int main() {
    IOS;
    rep (i, 0, 8) rep (j, 0, 8) cin >> a[i][j], --a[i][j];
    bool g = true;
    rep (i, 0, 8) {
        memset(f, 0, sizeof f);
        rep (j, 0, 8) {
            f[a[i][j]] |= 1, f[a[j][i]] |= 2;
            f[a[i / 3 * 3 + j / 3][i % 3 * 3 + j % 3]] |= 4;
        }
        rep (j, 0, 8) if (f[j] != 7) g = false;
    }
    cout << (g ? "Yes" : "No");
    return 0;
}

D - Good Tuple Problem

判断二分图

int h[N], to[M], ne[M], tot;
int a[M >> 1], color[N];

void add(int u, int v) {
    ne[++tot] = h[u];
    to[h[u] = tot] = v;
}

bool dfs(int x, int c) {
    color[x] = c;
    for (int i = h[x], y = to[i]; i; y = to[i = ne[i]]) {
        if (color[y] == c) return false;
        if (!color[y] && !dfs(y, -c)) return false;
    }
    return true;
}

int main() {
    IOS;
    cin >> n >> m;
    rep (i, 1, m) cin >> a[i];
    rep (i, 1, m) cin >> k, add(a[i], k), add(k, a[i]);
    bool f = true;
    rep (i, 1, n) {
        if (color[i]) continue;
        if (!dfs(i, 1)) { f = false; break; }
    }
    cout << (f ? "Yes" : "No");
    return 0;
}

E - Maximize Rating

dp

\(x_k = \textstyle \sum_{i=1}^{k} 0.9^i\)

$y_k = \frac{1200}{ \sqrt{k} } $

\(d_k = max(\textstyle \sum_{i=1}^{k} 0.9^i \times Q_i)\)

\(ans = max(\frac{d_k}{x_k} - y_k)\)

int a[N];
double x[N], y[N], d[N];

int main() {
    IOS;
    cin >> n; x[0] = 0, y[0] = 1;
    double ans = -1500, z = 1;
    rep (i, 1, n) {
        cin >> a[i], x[i] = z + x[i - 1], z *= 0.9, y[i] = 1200 / sqrt(i);
        d[i] = -1500;
        per (j, i, 1) {
            d[j] = max(d[j], d[j - 1] * 0.9 + a[i]);
            ans = max(ans, d[j] / x[j] - 1200 / y[j]);
        }
    }
    cout << precision(6) << ans;
    return 0;
}

F - Apples Editorial

扫描线

const int N = 2e5 + 5;

int n, m, _, k, cas;
vector<PII> st[N];

struct BIT {
    struct Node {
        int l, r, val, tag;
    } tr[N << 2];
    void push_up(int rt) {
        tr[rt].val = max(tr[rt << 1].val, tr[rt << 1 | 1].val);
    }
    void push_down(int rt) {
        tr[rt << 1].val += tr[rt].tag;
        tr[rt << 1 | 1].val += tr[rt].tag;
        tr[rt << 1].tag += tr[rt].tag;
        tr[rt << 1 | 1].tag += tr[rt].tag;
        tr[rt].tag = 0;
    }
    void build(int rt, int l, int r) {
        tr[rt].l = l, tr[rt].r = r;
        tr[rt].val = tr[rt].tag = 0;
        if (l == r) return;
        int mid = l + r >> 1;
        build(rt << 1, l, mid); build(rt << 1 | 1, mid + 1, r);
    }
    void change(int rt, int l, int r, int k) {
        if (tr[rt].l >= l && tr[rt].r <= r) { 
            tr[rt].val += k;
            tr[rt].tag += k;
            return;
        }
        push_down(rt);
        int mid = tr[rt].l + tr[rt].r >> 1;
        if (mid >= l) change(rt << 1, l, r, k);
        if (mid < r) change(rt << 1 | 1, l, r, k);
        push_up(rt);
    }
    int ask(int rt, int l, int r) {
        if (tr[rt].l >= l && tr[rt].r <= r) return tr[rt].val;
        push_down(rt);
        int mid = tr[rt].l + tr[rt].r >> 1;
        if (r <= mid) return ask(rt << 1, l, r);
        if (l > mid) return ask(rt << 1 | 1, l, r);
        return max(ask(rt << 1, l, r), ask(rt << 1 | 1, l, r));
    }
} bit;

int main() {
    IOS;
    int d, w; cin >> n >> d >> w;
    rep (i, 1, n) {
        int t, x; cin >> t >> x;
        st[max(1, t - d + 1)].emplace_back(x, 1);
        st[t + 1].emplace_back(x, -1);
    }
    bit.build(1, 1, N - 5);
    int ans = 0;
    for (int i = 1; i <= N - 5; ++i) {
        for (auto &p : st[i]) {
            bit.change(1, max(1, p.first - w + 1), p.first, p.second);
        }
        ans = max(ans, bit.ask(1, 1, N - 5));
    }
    cout << ans;
    return 0;
}