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。
【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
";
else if (cnt == 1)cout << "Same
";
else if (cnt >= 2)cout << "Even
";
}
【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
";
else if (n % 2 == 0)cout << "Same
";
else cout << "Odd
";
}
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) imes min(B))
先将序列按升序排序，
[egin{split}
ans &= sum_{i = 1}^{n}sum_{j = i + 1}^na_i imes a_j + sum_{i = 1}^na_i^2 \
&=sum_{i = 1}^na_i(sum_{j = i + 1}^na_j imes 2^{j - i + 1}） + sum_{i = 1}^na_i^2\
& 令 f(i) = sum_{j = i + 1}^na_j imes 2^{j - i + 1}\
&f(i - 1) = sum_{j = i + 1}^na_j imes 2^{j - i}\
& o f(i) = 2 imes 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 << "
";
}
C - Multiple Sequences
给定 (n(1le nle 2e5)) 和 (m(1le m le 2e5)) ，询问有多少满足长度为 (n) 的序列 (A)
- (1le 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) imes 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 << '
';
}
D - I Wanna Win The Game
给出 (n(1le n le5000)) 和 (m(1le 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 imes j) imes f[(i -2 imes 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。
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
", sum);
return 0;
}