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

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

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【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 \(N = 4k + r\)

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

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

  若 \(d|(2k + 1)\) 且 \(d\) 不是 \(1\) 和 \(2k + 1\)，则 \(d\) 一定为奇数，且同时会贡献 \(2d\) 这一偶数因子

• \(r = 0\) 时，\(N = 4k\)

  偶数因子个数至少是奇数因子的两倍

• \(r = 1\ or\ 3\) 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 \(\sum max(B)\times min(B)\)

---

先将序列按升序排序，

\[\begin{split}
ans &= \sum_{i = 1}^{n}\sum_{j = i + 1}^na_i\times a_j + \sum_{i = 1}^na_i^2 \\
&=\sum_{i = 1}^na_i(\sum_{j = i + 1}^na_j\times 2^{j - i + 1}） + \sum_{i = 1}^na_i^2\\
& 令 f(i) = \sum_{j = i + 1}^na_j\times 2^{j - i + 1}\\
&f(i - 1) = \sum_{j = i + 1}^na_j\times 2^{j - i}\\
&\to f(i) = 2\times f(i - 1) + a_i^2
\end{split}
\]

• 时间复杂度：\(\mathcal{O}(n)\)

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 \(n(1\le n\le 2e5)\) 和 \(m(1\le m \le 2e5)\) ，询问有多少满足长度为 \(n\) 的序列 \(A\)

• \(1\le A_i \le M(i = 1,2,...,N)\)
• \(A_{i + 1}\) 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 \(\mathcal{O}(nlog\ n)\)

先计算dp方案数。

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

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 \(n(1\le n \le5000)\) 和 \(m(1\le m \le5000)\) ，询问有多少满足条件的长度为 \(n\) 的序列 \(A\)。

• \(\sum_iA_i = m\)
• \(xor(A_i) = 0\)
• \(A_i>=0\)

---

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\((i−2×j)/2\) 的 \(n\) 个数左移一位，并选\(2×j\) 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 \(n(1≤n≤2e5)\) 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。

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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 \(g_u=mid\) 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;
}