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AtCoder Beginner Contest 332

发布时间 2023-12-10 22:15:50作者: 洛绫璃

AtCoder Beginner Contest 332

A - Online Shopping

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> n >> m >> k;
        ll ans = 0;
        rep (i, 1, n) {
            ll a, b; cin >> a >> b;
            ans += a * b;
        }
        ans += (ans >= m ? 0 : k);
        cout << ans;
    }
    return 0;
}

B - Glass and Mug

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> k >> n >> m;
        int a = 0, b = 0;
        rep (i, 1, k) {
            if (a == n) a = 0;
            else if (b == 0) b = m;
            else if (n - a > b) a += b, b = 0;
            else b -= n - a, a = n;
        }
        cout << a << ' ' << b;
    }
    return 0;
}

C - T-shirts

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> n >> m >> s;
        int ans = 0, a = m, b = 0;
        rep (c, 0, n) {
            bool f = 1;
            b = c; a = m;
            rep (i, 0, n - 1) {
                if (s[i] == '0') a = m, b = c;
                else if (s[i] == '1') {
                    if (a > 0) --a;
                    else if (b > 0) --b;
                    else { f = 0; break; }
                } else {
                    if (b > 0) --b;
                    else { f = 0; break; }
                }
            }
            if (f) { ans = c; break; }
        }
        cout << ans;
    }
    return 0;
}

D - Swapping Puzzle

int a[N][N], b[N][N];

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> n >> m;
        rep (i, 0, n - 1) rep (j, 0, m - 1) cin >> a[i][j];
        rep (i, 0, n - 1) rep (j, 0, m - 1) cin >> b[i][j];
        vector<int> x(n), z(m);
        rep (i, 0, n - 1) x[i] = i;
        rep (i, 0, m - 1) z[i] = i;
        int ans = 1e9;
        do {
            vector<int> y = z;
            do {
                bool f = 1;
                rep (i, 0, n - 1) rep (j, 0, m - 1) {
                    if (b[i][j] != a[x[i]][y[j]]) f = 0;
                }
                if (!f) continue;
                vector<int> z = x;
                int cnt = 0;
                rep (i, 0, n - 1) rep (j, i + 1, n - 1) if (z[i] > z[j]) ++cnt;
                z = y;
                rep (i, 0, m - 1) rep (j, i + 1, m - 1) if (z[i] > z[j]) ++cnt;
                ans = min(ans, cnt);
            } while (next_permutation(y.begin(), y.end()));
        } while (next_permutation(x.begin(), x.end()));
        cout << (ans == 1e9 ? -1 : ans);
    }
    return 0;
}

E - Lucky bag

看 t 神代码会的，这里需要一个证明，枚举子集的 dp 复杂度只有 \(3^n\)

一般模板长这样

for (int t = 0; t < 1 << n; ++t)
    for (int u = t; ; u = u - 1 & t) {
        // dp logic
        if (!u) break;
    }

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> n >> m;
        ll s = 0;
        rep (i, 0, n - 1) cin >> a[i], s += a[i];
        double avg = 1.0 * s / m;
        vector<double> cost(1 << n);
        for (int t = 0; t < 1 << n; ++t) {
            int cur = 0;
            rep (i, 0, n - 1) if (t >> i & 1) cur += a[i];
            cost[t] = (cur - avg) * (cur - avg);
        }

        const double inf = 1e30;
        vector<double> d(1 << n, inf); d[0] = 0;
        rep (_, 1, m) {
            vector<double> tmp(1 << n, inf);
            for (int t = 0; t < 1 << n; ++t)
                for (int u = t; ; u = u - 1 & t) {
                    umin(tmp[t], d[u] + cost[u ^ t]);
                    if (!u) break;
                }
            swap(tmp, d);
        }
        cout << precision(12) << d.back() / m;
    }
    return 0;
}

F - Random Update Query

推个式子，用线段树维护就好了，没有区间查询，我就没有写 push up 的逻辑

const int N = 2e5 + 5, mod = 998244353;

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

int qpow(int a, int b) {
    int ans = 1;
    for (; b; b >>= 1, a = 1ll * a * a % mod) if (b & 1) ans = 1ll * ans * a % mod;
    return ans;
}

struct BitTree {
    struct node { int l, r, val, tag_k, tag_x; } tr[N << 2];
    void push_down(int rt) {
        if (tr[rt << 1].l == tr[rt << 1].r) {
            tr[rt << 1].val = (1ll * tr[rt << 1].val * tr[rt].tag_k % mod + tr[rt].tag_x) % mod;
        } else {
            tr[rt << 1].tag_x = (1ll * tr[rt << 1].tag_x * tr[rt].tag_k % mod + tr[rt].tag_x) % mod;
            tr[rt << 1].tag_k = 1ll * tr[rt << 1].tag_k * tr[rt].tag_k % mod;
        }

        if (tr[rt << 1 | 1].l == tr[rt << 1 | 1].r) {
            tr[rt << 1 | 1].val = (1ll * tr[rt << 1 | 1].val * tr[rt].tag_k % mod + tr[rt].tag_x) % mod;
        } else {
            tr[rt << 1 | 1].tag_x = (1ll * tr[rt << 1 | 1].tag_x * tr[rt].tag_k % mod + tr[rt].tag_x) % mod;
            tr[rt << 1 | 1].tag_k = 1ll * tr[rt << 1 | 1].tag_k * tr[rt].tag_k % mod;
        }

        tr[rt].tag_x = 0, tr[rt].tag_k = 1;
    }
    void build(int rt, int l, int r, int* a) {
        tr[rt] = {l, r, 0, 1, 0};
        if (l == r) { tr[rt].val = a[l]; return; }
        int mid = l + r >> 1;
        build(rt << 1, l, mid, a); build(rt << 1 | 1, mid + 1, r, a);
    }
    void change(int rt, int l, int r, int k, int x) {
        if (r < l) return;
        if (tr[rt].l >= l && tr[rt].r <= r) {
            if (tr[rt].l == tr[rt].r) {
                tr[rt].val = (1ll * tr[rt].val * k % mod + x) % mod;
            } else {
                tr[rt].tag_x = (1ll * tr[rt].tag_x * k % mod + x) % mod;
                tr[rt].tag_k = 1ll * tr[rt].tag_k * k % mod;
            }
            return;
        }
        push_down(rt);
        int mid = tr[rt].l + tr[rt].r >> 1;
        if (l <= mid) change(rt << 1, l, r, k, x);
        if (r > mid) change(rt << 1 | 1, l, r, k, x);
    }
    ll ask(int rt, int x) {
        if (tr[rt].l == x && tr[rt].r == x) return tr[rt].val;
        push_down(rt);
        int mid = tr[rt].l + tr[rt].r >> 1;
        if (x <= mid) return ask(rt << 1, x);
        return ask(rt << 1 | 1, x);
    }
} bit;

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> n >> m;
        ll s = 0;
        rep (i, 1, n) cin >> a[i];
        bit.build(1, 1, n, a);
        rep (i, 1, m) {
            int l, r, t; cin >> l >> r >> t;
            int mm = qpow(r - l + 1, mod - 2);
            int x = 1ll * t * mm % mod, k = 1ll * (r - l) * mm % mod;
            bit.change(1, l, r, k, x);
        }
        rep (i, 1, n) cout << bit.ask(1, i) << ' ';
    }
    return 0;
}