TOYOTA SYSTEMS Programming Contest 2023(AtCoder Beginner Contest 330)

A - Counting Passes

int main() {
    IOS;
    cin >> n >> m;
    int ans = 0;
    rep (i, 1, n) cin >> k, ans += k >= m;
    cout << ans;
    return 0;
}

B - Minimize Abs 1

int main() {
    IOS;
    int l, r; cin >> n >> l >> r;
    rep (i, 1, n) {
        cin >> m;
        int y = m;
        if (r <= m) y = r;
        else if (l >= m) y = l;
        cout << y << ' ';
    }
    return 0;
}

C - Minimize Abs 2

考虑枚举 x，然后通过开根号贪心找 y

int main() {
    IOS;
    long long x; cin >> x;
    ll ans = x;
    for (ll i = 0, ii = i * i; ; ++i, ii = i * i) {
        ll s = x - ii;
        ans = min(ans, abs(s));
        if (s <= 0) break;
        ll j = ceil(sqrt(s));
        ans = min(ans, abs(j * j + ii - x));
        ans = min(ans, abs((j - 1) * (j - 1) + ii - x));
        if (j <= i) break;
    }
    cout << ans;
    return 0;
}

D - Counting Ls

枚举三角形的中心点 A，然后去枚举同 X 轴的点 B，再枚举同 y 轴的点 C。

然后就是推公式，化简。

int main() {
    IOS;
    cin >> n;
    ll ans = 0, cnt = 0;
    vector<vector<int>> a(n);
    vector<int> b(n);
    rep (i, 0, n - 1) rep (j, 0, n - 1) {
        char c; cin >> c;
        if (c == 'o') a[i].pb(j), ++b[j];
    }
    rep (i, 0, n - 1) {
        int sz = a[i].size();
        int c = 0;
        for (auto &x : a[i]) c += b[x] - 1;
        ans += 1ll * c * (sz - 1);
    }
    cout << ans;
    return 0;
}

E - Mex and Update

线段树维护 MEX, 只需要维护 [0, 20000] 区间即可。

const int N = 2e5 + 5;

int n, m, _, k, cas;
int a[N];

struct BitTree {
    struct node { int l, r, val; } tr[N << 2];
    void push_up(int rt) {
        tr[rt].val = 0;
        if (tr[rt << 1].l == tr[rt << 1].r) tr[rt].val += tr[rt << 1].val > 0;
        else tr[rt].val += tr[rt << 1].val;

        if (tr[rt << 1 | 1].l == tr[rt << 1 | 1].r) tr[rt].val += tr[rt << 1 | 1].val > 0;
        else tr[rt].val += tr[rt << 1 | 1].val;
    }
    void build(int rt, int l, int r) {
        tr[rt] = {l, r, 0};
        if (l == r) { tr[rt].val = 0; return; }
        int mid = l + r >> 1;
        build(rt << 1, l, mid); build(rt << 1 | 1, mid + 1, r);
    }
    void change(int rt, int x, int k) {
        if (tr[rt].l == x && tr[rt].r == x) {
            tr[rt].val += k;
            return;
        }
        int mid = tr[rt].l + tr[rt].r >> 1;
        if (x <= mid) change(rt << 1, x, k);
        else change(rt << 1 | 1, x, k);
        push_up(rt);
    }
    int ask(int rt) {
        if (tr[rt].l == tr[rt].r) return tr[rt].l;
        if (tr[rt << 1].val < tr[rt << 1].r - tr[rt << 1].l + 1) return ask(rt << 1);
        return ask(rt << 1 | 1);
    }
} bit;

int main() {
    IOS;
    bit.build(1, 0, 200001);
    cin >> n >> m;
    rep (i, 1, n) {
        cin >> a[i];
        if (a[i] <= 200000)
            bit.change(1, a[i], 1);
    }
    rep (i, 1, m) {
        int x, y; cin >> x >> y;
        if (a[x] <= 200000) bit.change(1, a[x], -1);
        a[x] = y;
        if (a[x] <= 200000) bit.change(1, a[x], 1);
        cout << bit.ask(1) << '\n';
    }
    return 0;
}

F - Minimize Bounding Square

倍增，把 x 轴 y 轴分开处理即可

const int N = 2e5 + 5;

int n, m, _, cas;
ll k, sx[N], sy[N];
int x[N], y[N];

ll check(int d, ll s[], int a[]) {
    ll mi = 1ll << 60;
    rep (i, 2, n * 2 + 1) {
        int c = a[i >> 1] - (i & 1 ? d : 0);
        ll res = 0;
        int x = lower_bound(a + 1, a + 1 + n, c) - a - 1;
        res += 1ll * c * x - s[x];
        x = lower_bound(a + 1, a + 1 + n, c + d) - a - 1;
        res += s[n] - s[x] - 1ll * (c + d) * (n - x);
        umin(mi, res);
    }
    return mi;
}

int main() {
    IOS;
    cin >> n >> k;
    rep (i, 1, n) cin >> x[i] >> y[i];
    sort(x + 1, x + 1 + n); sort(y + 1, y + 1 + n);
    rep (i, 1, n) sx[i] = sx[i - 1] + x[i], sy[i] = sy[i - 1] + y[i];
    int ans = (1 << 30) - 1;
    per (i, 29, 0) if (check(ans - (1 << i), sx, x) + check(ans - (1 << i), sy, y) <= k)
        ans -= 1 << i;
    cout << ans;
    return 0;
}