AtCoder Regular Contest 116 (A~F补题记录）

补题链接：Here

第一次打 ARC，被数学题虐惨了

赛后部分数学证明学习自 ACwisher

A - Odd vs Even

$T (1 \leq T \leq 2 \times 10^{5})$ 组测试数据，每次询问一个正整数 $N (1 \leq N \leq 2 \times 10^{18})$ 的奇数因子多还是偶数因子多。

【方案一】

设n有cnt个质因子2,
假设n有x个奇数因子,那么就会有m*(2^(cnt)-1)种偶数因子,即用cnt个2的子集和奇数因子配对.

因此:
当cnt=0时,没有偶数因子,此时奇数因子多.
当cnt=1时,偶数因子和奇数因子一样多.
当cnt>=2时,偶数因子比奇数因子多.

using ll = long long;
void solve() {
    ll cnt = 0, n; cin >> n;
    while (n % 2 == 0) n /= 2, cnt++;
    if (cnt == 0)cout << "Odd\n";
    else if (cnt == 1)cout << "Same\n";
    else if (cnt >= 2)cout << "Even\n";
}

【方案二】

赛后打了一下表，

发现设 $N = 4 k + r$

$r = 2$ 时， $N = 4 k + 2 = 2 (2 k + 1)$

偶数因子有 $2$ 和 $2 (2 k + 1)$ ，奇数因子有 $1$ 和 $2 k + 1$

若 $d | (2 k + 1)$ 且 $d$ 不是 $1$ 和 $2 k + 1$ ，则 $d$ 一定为奇数，且同时会贡献 $2 d$ 这一偶数因子
$r = 0$ 时， $N = 4 k$

偶数因子个数至少是奇数因子的两倍
$r = 1 o r 3$ 是

偶数因子个数为 0 个，奇数因子至少 2 个

using ll = long long;
void solve() {
    ll n; cin >> n;
    if (n % 4 == 0)cout << "Even\n";
    else if (n % 2 == 0)cout << "Same\n";
    else cout << "Odd\n";
}

B - Products of Min-Max

给出一个包含 $n$ 个数的序列 $A$ ，有 $2^{n - 1}$ 个 $A$ 的非空子序列 $B$

求 $\sum m a x (B) \times m i n (B)$

先将序列按升序排序，

\begin{aligned} a n s & = \sum_{i = 1}^{n} \sum_{j = i + 1}^{n} a_{i} \times a_{j} + \sum_{i = 1}^{n} a_{i}^{2} \\ = \sum_{i = 1}^{n} a_{i} (\sum_{j = i + 1}^{n} a_{j} \times 2^{j - i + 1} ） + \sum_{i = 1}^{n} a_{i}^{2} \\ 令 f (i) = \sum_{j = i + 1}^{n} a_{j} \times 2^{j - i + 1} \\ f (i - 1) = \sum_{j = i + 1}^{n} a_{j} \times 2^{j - i} \\ \to f (i) = 2 \times f (i - 1) + a_{i}^{2} \end{aligned}

时间复杂度： $O (n)$

using ll = long long;
const int N = 2e5 + 10, mod =  998244353;
ll a[N], n;
void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i)cin >> a[i];
    sort(a + 1, a + 1 + n);
    ll ans = 0;
    for (int i = n, tmp = 0; i >= 1; --i) {
        ans = (ans + a[i] * a[i] % mod) % mod;
        ans  = (ans + a[i] * tmp % mod) % mod;
        tmp = (2ll * tmp + a[i]) % mod;
    }
    cout << ans << "\n";
}

C - Multiple Sequences

给定 $n (1 \leq n \leq 2 e 5)$ 和 $m (1 \leq m \leq 2 e 5)$ ，询问有多少满足长度为 $n$ 的序列 $A$

$1 \leq A_{i} \leq M (i = 1, 2, . . ., N)$
$A_{i + 1}$ 是 $A_{i}$ 的倍数 $(i = 1, 2, . . ., N - 1)$

注意到如果每次都有改变，顶多有 $19$ 个数。调和级数一下是 $O (n l o g n)$

先计算dp方案数。

枚举有 $i$ 个不同的数，对答案的贡献为 $C (n - 1, i - 1) \times \sum_{j = 1}^{m} d p [i] [j]$

注意第一个肯定是第一个，无需考虑

using ll = long long;
const int N = 2e5 + 10, mod = 998244353, K = 25;
int n, m, f[K][N], fac[N], ifac[N];
ll qpow(int x, int y ) {
    ll ans = 1;
    for (; y; y >>= 1, x = 1ll * x * x % mod)
        if (y & 1) ans = ans * x % mod;
    return ans;
}
int C(int x, int y) {
    return 1ll * fac[x] * ifac[y] % mod * ifac[x - y] % mod;
}
void solve() {
    cin >> n >> m;
    for (int i = 1; i <= m; i++) f[1][i] = 1;
    for (int i = 2; i <= 19; i++)
        for (int j = 1; j <= m; j++)
            for (int k = 2; 1ll * j * k <= m; k++)
                f[i][j * k] = (f[i][j * k] + f[i - 1][j]) % mod;
    fac[0] = ifac[0] = 1;
    for (int i = 1; i <= n; ++i)fac[i] = 1ll * fac[i - 1] * i % mod;
    ifac[n] = qpow(fac[n], mod - 2);
    for (int i = n - 1; i >= 1; i--) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
    // -------------------上方为初始化--------------------- //
    int ans = 0;
    for (int i = 1; i <= min(19, n); ++i) {
        int cnt = 0;
        for (int j = 1; j <= m; ++j)cnt = (cnt + f[i][j]) % mod;
        ans = (ans + 1ll * cnt * C(n - 1, i - 1) % mod) % mod;
    }
    cout << ans << '\n';
}

D - I Wanna Win The Game

给出 $n (1 \leq n \leq 5000)$ 和 $m (1 \leq m \leq 5000)$ ，询问有多少满足条件的长度为 $n$ 的序列 $A$ 。

$\sum_{i} A_{i} = m$
$x o r (A_{i}) = 0$
$A_{i} >= 0$

D题开始，没有做出来，参考了高 Rank 的解法

显然每一位都有偶数个数选择。

$f [i]$ 表示 $n$ 个数和为 $i$ 的方案数

$f [i] + = C (n, 2 \times j) \times f [(i - 2 \times j) / 2]$

是将之前和为 $(i - 2 \times j) / 2$ 的 $n$ 个数左移一位，并选 $2 \times j$ 个数最后一位为 $1$

using ll = long long;
const int N = 5e3 + 10, mod = 998244353;
ll fac[N], ifac[N], f[N];
int n, m;
ll qpow(int x, int y ) {
    ll ans = 1;
    for (; y; y >>= 1, x = 1ll * x * x % mod)
        if (y & 1) ans = ans * x % mod;
    return ans % mod;
}
ll C(int x, int y) {
    return 1ll * fac[x] * ifac[y] % mod * ifac[x - y] % mod;
}
void solve() {
    cin >> n >> m;
    fac[0] = ifac[0] = 1;
    for (int i = 1; i <= n; ++i)fac[i] = 1ll * fac[i - 1] * i % mod;
    ifac[n] = qpow(fac[n], mod - 2);
    for (int i = n -  1; i >= 1; i--)ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
    // ----------------------------------- //
    f[0] = 1;
    for (int i = 1; i <= m; ++i) {
        if (i & 1)continue;
        for (int j = 0; j <= m and i - 2 * j >= 0; ++j)
            f[i] = (f[i] + 1ll * C(n, 2 * j) * f[(i - 2 * j) / 2] % mod ) % mod;
    }
    cout << f[m];
}

E - Spread of Information

有 $n (1 \leq n \leq 2 e 5)$ 个点，选择 $k$ 个为初始感染点，每秒沿边传播（扩张），求最快时间。

二分最快时间（距离）

$f_{u} u$ 子树内离他最近的感染点的距离
$g_{u} u$ 子树内离他最远的非感染点的距离

如果通过根节点中转能帮上 $g_{u}$ 那么一定整个子树都已被覆盖

其他情况，如果 $g_{u} = m i d$ 那 $u$ 必须成为初始感染点

using ll = long long;
const int N = 2e5 + 10, inf = 0x3f3f3f3f;
vector<int>e[N];
int n, k, mid, ret;
int f[N], g[N];
void dfs(int u, int fa) {
    g[u] = 0, f[u] = inf;
    for (int v : e[u]) {
        if (v == fa)continue;
        dfs(v, u);
        f[u] = min(f[u], f[v] + 1);
        g[u] = max(g[u], g[v] + 1);
    }
    if (f[u] + g[u] <= mid) g[u] = -inf;
    else if (g[u] == mid) f[u] = 0, g[u] = -inf, ret++;
}
bool check() {
    ret = 0;
    dfs(1, 0);
    if (g[1] >= 0)ret++;
    return ret <= k;
}
void solve() {
    cin >> n >> k;
    for (int i = 1, u, v; i < n; ++i) {
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    int l = 0, r =  n, ans = n;
    while (l <= r) {
        mid = (r + l) >> 1;
        if (check())r = mid - 1, ans = mid;
        else l = mid + 1;
    }
    cout << ans;
}

F - Deque Game

const int N = 2e5 + 10;
int n, q[N];
vector<int> a[N];
int main() {
    scanf("%d", &n);
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        int k; scanf("%d", &k);
        for (int j = 0, x; j < k; j++)
            scanf("%d", &x), a[i].push_back(x);
        cnt += (k & 1 ^ 1);
    }
    ll sum = 0; int len = 0;
    for (int i = 1; i <= n; i++) {
        if (a[i].size() & 1 ^ 1) {
            if (a[i].size() == 2) {
                sum += min(a[i][0], a[i][1]);
                q[++len] = - (max(a[i][0], a[i][1]) - min(a[i][0], a[i][1]));
            } else {
                int mid0 = a[i].size() / 2 - 1, mid1 = a[i].size() / 2 + 1 - 1, ret0, ret1;
                if (cnt & 1 ^ 1) {
                    ret0 = min(a[i][mid0], max(a[i][mid0 - 1], a[i][mid0 + 1]));
                    ret1 = min(a[i][mid1], max(a[i][mid1 - 1], a[i][mid1 + 1]));
                } else {
                    ret0 = max(a[i][mid0], min(a[i][mid0 - 1], a[i][mid0 + 1]));
                    ret1 = max(a[i][mid1], min(a[i][mid1 - 1], a[i][mid1 + 1]));
                }
                sum += min(ret0, ret1);
                q[++len] = - (max(ret0, ret1) - min(ret0, ret1));
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        if (a[i].size() & 1) {
            if (a[i].size() == 1) {
                sum += a[i][0];
            } else {
                int mid0 = (a[i].size() + 1) / 2 - 1;
                if (cnt & 1 ^ 1)
                    sum += min(a[i][mid0], max(a[i][mid0 - 1], a[i][mid0 + 1]));
                else
                    sum += max(a[i][mid0], min(a[i][mid0 - 1], a[i][mid0 + 1]));
            }
        }
    }
    sort(q + 1, q + len + 1);
    for (int i = 1; i <= len; i += 2) sum -= q[i];
    printf("%lld\n", sum);
    return 0;
}