标签：std AtCoder return Beginner int poly 267 const size
E - Erasing Vertices 2
做法1
观察可得：对于某个时刻，贪心删当前代价最小的点肯定是最优的。
但是删一个点会减少相邻接的点的代价。然后就想到了堆，但是这个堆需要支持decrease-key操作。
decrease-key 这个操作std::priority_queue并不支持，但是其实二叉堆也能做到 \(O(\log n)\)。
pbds 里的堆支持decrease-key，接口为 modify，然后就完事了。
做法2
这个是官方题解给出的做法，大概就是二分答案，然后通过搜索把代价小于等于当前二分的值的点都删掉，如果所有点都能删掉说明当前二分的代价可行。
AC代码
// Problem: E - Erasing Vertices 2
// Contest: AtCoder - NEC Programming Contest 2022 (AtCoder Beginner Contest 267)
// URL: https://atcoder.jp/contests/abc267/tasks/abc267_e
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define CPPIO std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
#ifdef BACKLIGHT
#include "debug.h"
#else
#define logd(...) ;
#define ASSERT(x) ;
#define serialize() ""
#endif
using i64 = int64_t;
using u64 = uint64_t;
void Initialize();
void SolveCase(int Case);
int main(int argc, char* argv[]) {
CPPIO;
int T = 1;
// std::cin >> T;
for (int t = 1; t <= T; ++t) {
SolveCase(t);
}
return 0;
}
void Initialize() {}
#include <ext/pb_ds/priority_queue.hpp>
template <typename T>
using min_heap = __gnu_pbds::priority_queue<T, std::greater<T>, __gnu_pbds::pairing_heap_tag>;
void SolveCase(int Case) {
int n, m;
std::cin >> n >> m;
std::vector<int> a(n);
for (int i = 0; i < n; ++i) {
std::cin >> a[i];
}
std::vector<std::vector<int>> g(n);
for (int i = 0; i < m; ++i) {
int u, v;
std::cin >> u >> v;
--u, --v;
g[u].push_back(v);
g[v].push_back(u);
}
std::vector<i64> s(n);
for (int i = 0; i < n; ++i) {
for (int v : g[i])
s[v] += a[i];
}
min_heap<std::pair<i64, int>> q;
std::vector<min_heap<std::pair<i64, int>>::point_iterator> p(n);
for (int i = 0; i < n; ++i) {
p[i] = q.push({s[i], i});
}
i64 ans = 0;
while (!q.empty()) {
auto [cost, id] = q.top();
q.pop();
ans = std::max(ans, cost);
p[id] = nullptr;
for (int v : g[id]) {
if (p[v] == nullptr)
continue;
auto [first, second] = *p[v];
q.modify(p[v], {first - a[id], second});
}
}
std::cout << ans << "\n";
}
F - Exactly K Steps
有个结论：假设树的直径的两个端点分别为 \(a\) 和 \(b\)，那么对于树中任意一点 \(u\) ，\(a\) 和 \(b\) 中之一必定满足其是树中所有点中离\(u\)距离最大的点。用反证法可以证明。
由此，对于某个询问\(u, k\)，要么能在 \(u\) 到 \(a\) 或 \(u\) 到 \(b\) 的路径中找到答案，要么无解。
做法1
想着把直径端点之一搞成根，此时直径是一条一端为根一端为叶子的链。
对于一个点，往父亲的方向条一定会跳到这条链上。
然后先跳到这条链上，假设此时位于\(u\)，那么\(u\)所在的长链一定是直径，这个时候再枚举往父亲的方向跳，还是往长链的方向条，就能枚举到\(u\) 到 \(a\) 和 \(u\) 到 \(b\) 的路径。
往父亲的方向跳和往长链的方向条都可以用倍增加速。
做法2
这个是官方题解给出的做法，大概就是离线然后分别以 \(a\) 和 \(b\) 作为根，一共跑两次，这样就只需要考虑往父亲的方向跳的情况，也能枚举到\(u\) 到 \(a\) 和 \(u\) 到 \(b\) 的路径，而且相对做法1会简单一点。
AC代码
// Problem: F - Exactly K Steps
// Contest: AtCoder - NEC Programming Contest 2022 (AtCoder Beginner Contest 267)
// URL: https://atcoder.jp/contests/abc267/tasks/abc267_f
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define CPPIO std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
#ifdef BACKLIGHT
#include "debug.h"
#else
#define logd(...) ;
#define ASSERT(x) ;
#define serialize() ""
#endif
using i64 = int64_t;
using u64 = uint64_t;
void Initialize();
void SolveCase(int Case);
int main(int argc, char* argv[]) {
CPPIO;
int T = 1;
// std::cin >> T;
for (int t = 1; t <= T; ++t) {
SolveCase(t);
}
return 0;
}
void Initialize() {}
void SolveCase(int Case) {
int n, m = 20;
std::cin >> n;
std::vector<std::vector<int>> g(n);
for (int i = 0; i < n - 1; ++i) {
int u, v;
std::cin >> u >> v;
--u, --v;
g[u].push_back(v);
g[v].push_back(u);
}
// make one end of diameter the root.
std::vector<int> dep(n), f(n);
int rt = 0;
std::function<void(int, int, int)> dfs1 = [&](int u, int fa, int depth) {
dep[u] = depth;
f[u] = fa;
if (dep[u] > dep[rt])
rt = u;
for (int v : g[u]) {
if (v == fa)
continue;
dfs1(v, u, depth + 1);
}
};
dfs1(0, 0, 0);
int temp = rt;
dfs1(rt, rt, 0);
rt = temp;
logd(rt);
logd(dep);
// binary lifting for k-th ancester.
std::vector<std::vector<int>> up(n, std::vector<int>(m));
std::function<void(int, int)> dfs2 = [&](int u, int fa) {
if (u == rt) {
std::fill(up[u].begin(), up[u].end(), rt);
} else {
up[u][0] = fa;
for (int i = 1; i < m; ++i) {
up[u][i] = up[up[u][i - 1]][i - 1];
}
}
for (int v : g[u]) {
if (v == fa)
continue;
dfs2(v, u);
}
};
dfs2(rt, rt);
logd(up);
// binary lifting for longest chain in subtree.
std::vector<int> len(n), son(n);
std::vector<std::vector<int>> down(n, std::vector<int>(m));
std::function<void(int, int)> dfs3 = [&](int u, int fa) {
son[u] = -1;
len[u] = 0;
for (int v : g[u]) {
if (v == fa)
continue;
dfs3(v, u);
if (son[u] == -1 || len[v] > len[son[u]]) {
son[u] = v;
len[u] = len[v] + 1;
}
}
if (son[u] == -1) {
std::fill(down[u].begin(), down[u].end(), u);
} else {
down[u][0] = son[u];
for (int i = 1; i < m; ++i) {
down[u][i] = down[down[u][i - 1]][i - 1];
}
}
};
dfs3(rt, rt);
std::vector<int> in_diameter(n, false);
{
int x = rt;
while (x != -1) {
in_diameter[x] = true;
x = son[x];
}
}
// return v whose distance from u is k by:
// - jump up in the path to root.
// - jump down in the longest chain in subtree whose root is u.
auto Q = [&](int u, int k) {
if (dep[u] >= k) {
int x = u;
for (int i = m - 1; i >= 0; --i) {
if ((k >> i) & 1)
x = up[x][i];
}
return x + 1;
}
if (len[u] >= k) {
int x = u;
for (int i = m - 1; i >= 0; --i) {
if ((k >> i) & 1)
x = down[x][i];
}
return x + 1;
}
return -1;
};
int q;
std::cin >> q;
for (int _ = 0; _ < q; ++_) {
int u, k, r;
std::cin >> u >> k;
--u;
r = Q(u, k);
if (r != -1) {
std::cout << r << "\n";
continue;
}
if (in_diameter[u]) {
std::cout << "-1\n";
continue;
}
int y = u;
for (int i = m - 1; i >= 0; --i) {
if (not in_diameter[up[y][i]])
y = up[y][i];
}
y = f[y];
k -= dep[u] - dep[y];
u = y;
std::cout << Q(u, k) << "\n";
}
}
G - Increasing K Times
问题转换成 \(a\) 有多少个排列 \(A\) 满足恰好有 \(k\) 个 \(i\) 满足\(A_i < A_{i + 1}\)。
考虑从小到大把 \(a\) 中的元素加入 \(A\) ，从而构成一个 \(a\) 的排列 \(A\) 。为了防止边界条件的讨论，不妨令初始时 \(A = [0, 0]\)，然后再加入元素。这样相比原本的情况中会多算一个满足\(A_i < A_{i + 1}\) 的 \(i\)，因为 \(A_n\) 一定大于 \(0\) ，此时改成求恰好有 \(k + 1\) 个满足条件的\(i\)即可。
然后考虑动态规划，令 \(dp_{i, j}\) 表示插入前 \(i\) 个元素，恰好有 \(j\) 个 \(x\) 满足 \(A_x < A_{x + 1}\)。
假设现在要插入\(y\)，这个操作要么令满足条件的 \(x\) 的数量保持不变，要么令满足条件的 \(x\) 的数量增加一。
什么时候会增加呢，只有将 \(y\) 插到满足 \(A_{x} \ge A_{x + 1}\) 的 \(A_x\) 之后才会令满足条件的 \(x\) 的数量增加。假设之前已经插入了 \(i\) 个元素，有 \(j\) 个满足条件的 \(x\)，然后插入了 \(z\) 个 \(y\)，那么一共有 \(i - j - z\) 个位置，将 \(y\) 插到这个位置会使满足条件的 \(x\) 数量增加一，剩余 \(j + z\) 个位置会使满足条件的 \(x\) 数量保持不变。
AC代码
// Problem: G - Increasing K Times
// Contest: AtCoder - NEC Programming Contest 2022 (AtCoder Beginner Contest 267)
// URL: https://atcoder.jp/contests/abc267/tasks/abc267_g
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define CPPIO std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
#ifdef BACKLIGHT
#include "debug.h"
#else
#define logd(...) ;
#define ASSERT(x) ;
#define serialize() ""
#endif
using i64 = int64_t;
using u64 = uint64_t;
void Initialize();
void SolveCase(int Case);
int main(int argc, char* argv[]) {
CPPIO;
int T = 1;
// std::cin >> T;
for (int t = 1; t <= T; ++t) {
SolveCase(t);
}
return 0;
}
void Initialize() {}
template <typename ValueType, ValueType mod_, typename SupperType = int64_t>
class Modular {
private:
ValueType value_;
ValueType normalize(ValueType value) const {
if (value >= 0 && value < mod_)
return value;
value %= mod_;
if (value < 0)
value += mod_;
return value;
}
ValueType power(ValueType value, size_t exponent) const {
ValueType result = 1;
ValueType base = value;
while (exponent) {
if (exponent & 1)
result = SupperType(result) * base % mod_;
base = SupperType(base) * base % mod_;
exponent >>= 1;
}
return result;
}
public:
Modular() : value_(0) {}
Modular(const ValueType& value) : value_(normalize(value)) {}
ValueType value() const { return value_; }
Modular inv() const { return Modular(power(value_, mod_ - 2)); }
Modular power(size_t exponent) const { return Modular(power(value_, exponent)); }
friend Modular operator+(const Modular& lhs, const Modular& rhs) {
ValueType result = lhs.value() + rhs.value() >= mod_ ? lhs.value() + rhs.value() - mod_
: lhs.value() + rhs.value();
return Modular(result);
}
friend Modular operator-(const Modular& lhs, const Modular& rhs) {
ValueType result = lhs.value() - rhs.value() < 0 ? lhs.value() - rhs.value() + mod_
: lhs.value() - rhs.value();
return Modular(result);
}
friend Modular operator*(const Modular& lhs, const Modular& rhs) {
ValueType result = SupperType(1) * lhs.value() * rhs.value() % mod_;
return Modular(result);
}
friend Modular operator/(const Modular& lhs, const Modular& rhs) {
ValueType result = SupperType(1) * lhs.value() * rhs.inv().value() % mod_;
return Modular(result);
}
};
template <typename StreamType, typename ValueType, ValueType mod, typename SupperType = int64_t>
StreamType& operator<<(StreamType& out, const Modular<ValueType, mod, SupperType>& modular) {
return out << modular.value();
}
template <typename StreamType, typename ValueType, ValueType mod, typename SupperType = int64_t>
StreamType& operator>>(StreamType& in, Modular<ValueType, mod, SupperType>& modular) {
ValueType value;
in >> value;
modular = Modular<ValueType, mod, SupperType>(value);
return in;
}
// using Mint = Modular<int, 1'000'000'007>;
using Mint = Modular<int, 998'244'353>;
class Binom {
private:
std::vector<Mint> f, g;
public:
Binom(int n) {
f.resize(n + 1);
g.resize(n + 1);
f[0] = Mint(1);
for (int i = 1; i <= n; ++i)
f[i] = f[i - 1] * Mint(i);
g[n] = f[n].inv();
for (int i = n - 1; i >= 0; --i)
g[i] = g[i + 1] * Mint(i + 1);
}
Mint operator()(int n, int m) {
if (n < 0 || m < 0 || m > n)
return Mint(0);
return f[n] * g[m] * g[n - m];
}
};
void SolveCase(int Case) {
int n, k;
std::cin >> n >> k;
++k;
std::vector<int> a(n + 1);
for (int i = 1; i <= n; ++i)
std::cin >> a[i];
std::sort(a.begin() + 1, a.end());
std::vector<int> same(n + 1);
for (int i = 1; i <= n; ++i) {
if (a[i] != a[i - 1])
same[i] = 0;
else
same[i] = same[i - 1] + 1;
}
std::vector<Mint> dp(k + 1, Mint(0));
dp[0] = Mint(1);
for (int i = 1; i <= n; ++i) {
std::vector<Mint> temp(k + 1, Mint(0));
for (int j = 0; j <= k; ++j) {
temp[j] = temp[j] + Mint(j + same[i]) * dp[j];
if (j + 1 <= k)
temp[j + 1] = temp[j + 1] + Mint(i - j - same[i]) * dp[j];
}
dp = temp;
}
std::cout << dp[k].value() << "\n";
}
Ex - Odd Sum
考虑动态规划，令 \(dp_{i, 0/1}\) 表示选奇数或者偶数个元素且所选元素之和为 \(i\) 的方案数，然后类似背包的DP就能求，但是复杂度爆炸。
考虑 \(f_{0}(x) = \sum_i a_i x^i\) ，\([x^i]f_0 (x)\)表示选偶数个元素且元素和为 \(i\) 的方案数，类似可得\(f_{1}(x)\)。
假设现在有两个数组，对应的多项式分别为\(f_{0/1}\)和\(g_{0/1}\)，那么对于两个数组的并集对应的多项式\(h\)有：
\[h_0 = f_0 * g_0 + f_1 * g_1\\ h_1 = f_0 * g_1 + f_1 * g_0\\ \]把每个 \(a_i\) 都看成独立的数组，不断两两合并的到 \(a\)，对应的多项式也两两合并，得到 \(a\) 对应的多项式\(f_{0/1}\)，答案即为\([x^m]f_1\)。
用类似分治FFT的过程跑能做到\(O(m \log m \log n)\)的时间复杂度。
AC代码
// Problem: Ex - Odd Sum
// Contest: AtCoder - NEC Programming Contest 2022 (AtCoder Beginner Contest 267)
// URL: https://atcoder.jp/contests/abc267/tasks/abc267_h
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define CPPIO std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
#ifdef BACKLIGHT
#include "debug.h"
#else
#define logd(...) ;
#define ASSERT(x) ;
#define serialize() ""
#endif
using i64 = int64_t;
using u64 = uint64_t;
void Initialize();
void SolveCase(int Case);
int main(int argc, char* argv[]) {
CPPIO;
int T = 1;
// std::cin >> T;
for (int t = 1; t <= T; ++t) {
SolveCase(t);
}
return 0;
}
void Initialize() {}
namespace Polynomial {
constexpr int P = 998244353, G = 3;
std::vector<int> rev, roots{0, 1};
int power(int a, int b) {
int r = 1;
while (b) {
if (b & 1)
r = 1ll * r * a % P;
a = 1ll * a * a % P;
b >>= 1;
}
return r;
}
void dft(std::vector<int>& a) {
int n = a.size();
if (int(rev.size()) != n) {
int k = __builtin_ctz(n) - 1;
rev.resize(n);
for (int i = 0; i < n; ++i)
rev[i] = rev[i >> 1] >> 1 | (i & 1) << k;
}
for (int i = 0; i < n; ++i)
if (rev[i] < i)
std::swap(a[i], a[rev[i]]);
if (int(roots.size()) < n) {
int k = __builtin_ctz(roots.size());
roots.resize(n);
while ((1 << k) < n) {
int e = power(G, (P - 1) >> (k + 1));
for (int i = 1 << (k - 1); i < (1 << k); ++i) {
roots[2 * i] = roots[i];
roots[2 * i + 1] = 1ll * roots[i] * e % P;
}
++k;
}
}
for (int k = 1; k < n; k *= 2) {
for (int i = 0; i < n; i += 2 * k) {
for (int j = 0; j < k; ++j) {
int u = a[i + j];
int v = 1ll * a[i + j + k] * roots[k + j] % P;
int x = u + v;
if (x >= P)
x -= P;
a[i + j] = x;
x = u - v;
if (x < 0)
x += P;
a[i + j + k] = x;
}
}
}
}
void idft(std::vector<int>& a) {
int n = a.size();
std::reverse(a.begin() + 1, a.end());
dft(a);
int inv = power(n, P - 2);
for (int i = 0; i < n; ++i)
a[i] = 1ll * a[i] * inv % P;
}
struct poly {
std::vector<int> a;
poly() {}
poly(int f0) { a = {f0}; }
poly(const std::vector<int>& f) : a(f) {
while (!a.empty() && !a.back())
a.pop_back();
}
poly(const std::vector<int>& f, int n) : a(f) { a.resize(n); }
int size() const { return a.size(); }
int deg() const { return a.size() - 1; }
int operator[](int idx) const {
if (idx < 0 || idx >= size())
return 0;
return a[idx];
}
std::string to_string() const {
std::stringstream ss;
ss << "poly: ";
for (int v : a)
ss << v << " ";
return ss.str();
}
poly mulxk(int k) const {
auto b = a;
b.insert(b.begin(), k, 0);
return poly(b);
}
poly modxk(int k) const {
k = std::min(k, size());
return poly(std::vector<int>(a.begin(), a.begin() + k));
}
poly alignxk(int k) const { return poly(a, k); }
poly divxk(int k) const {
if (size() <= k)
return poly();
return poly(std::vector<int>(a.begin() + k, a.end()));
}
friend poly operator+(const poly& f, const poly& g) {
int k = std::max(f.size(), g.size());
std::vector<int> res(k);
for (int i = 0; i < k; ++i) {
res[i] = f[i] + g[i];
if (res[i] >= P)
res[i] -= P;
}
return poly(res);
}
friend poly operator-(const poly& f, const poly& g) {
int k = std::max(f.size(), g.size());
std::vector<int> res(k);
for (int i = 0; i < k; ++i) {
res[i] = f[i] - g[i];
if (res[i] < 0)
res[i] += P;
}
return poly(res);
}
friend poly operator*(const poly& f, const poly& g) {
int sz = 1, k = f.size() + g.size() - 1;
while (sz < k)
sz *= 2;
std::vector<int> p = f.a, q = g.a;
p.resize(sz);
q.resize(sz);
dft(p);
dft(q);
for (int i = 0; i < sz; ++i)
p[i] = 1ll * p[i] * q[i] % P;
idft(p);
return poly(p);
}
friend poly operator/(const poly& f, const poly& g) { return f.divide(g).first; }
friend poly operator%(const poly& f, const poly& g) { return f.divide(g).second; }
poly& operator+=(const poly& f) { return (*this) = (*this) + f; }
poly& operator-=(const poly& f) { return (*this) = (*this) - f; }
poly& operator*=(const poly& f) { return (*this) = (*this) * f; }
poly& operator/=(const poly& f) { return (*this) = divide(f).first; }
poly& operator%=(const poly& f) { return (*this) = divide(f).second; }
poly derivative() const {
if (a.empty())
return poly();
int n = a.size();
std::vector<int> res(n - 1);
for (int i = 0; i < n - 1; ++i)
res[i] = 1ll * (i + 1) * a[i + 1] % P;
return poly(res);
}
poly integral() const {
if (a.empty())
return poly();
int n = a.size();
std::vector<int> res(n + 1);
for (int i = 0; i < n; ++i)
res[i + 1] = 1ll * a[i] * power(i + 1, P - 2) % P;
return poly(res);
}
poly rev() const { return poly(std::vector<int>(a.rbegin(), a.rend())); }
poly inv(int m) const {
poly x(power(a[0], P - 2));
int k = 1;
while (k < m) {
k *= 2;
x = (x * (2 - modxk(k) * x)).modxk(k);
}
return x.modxk(m);
}
poly log(int m) const { return (derivative() * inv(m)).integral().modxk(m); }
poly exp(int m) const {
poly x(1);
int k = 1;
while (k < m) {
k *= 2;
x = (x * (1 - x.log(k) + modxk(k))).modxk(k);
}
return x.modxk(m);
}
poly sqrt(int m) const {
poly x(1);
int k = 1;
while (k < m) {
k *= 2;
x = (x + (modxk(k) * x.inv(k)).modxk(k)) * ((P + 1) / 2);
}
return x.modxk(m);
}
poly sin() const {
int i = power(G, (P - 1) / 4);
poly p = i * (*this);
p = p.exp(p.size());
poly q = (P - i) * (*this);
q = q.exp(q.size());
poly r = (p - q) * power(2 * i % P, P - 2);
return r;
}
poly cos() const {
int i = power(G, (P - 1) / 4);
poly p = i * (*this);
p = p.exp(p.size());
poly q = (P - i) * (*this);
q = q.exp(q.size());
poly r = (p + q) * power(2, P - 2);
return r;
}
poly tan() const { return sin() / cos(); }
poly cot() const { return cos() / sin(); }
poly arcsin() {
poly sq = (*this) * (*this).modxk(size());
for (int i = 0; i < size(); ++i)
sq.a[i] = sq.a[i] ? P - sq.a[i] : 0;
sq.a[0] = 1 + sq.a[0];
if (sq.a[0] >= P)
sq.a[0] -= P;
poly r = (derivative() * sq.sqrt(size()).inv(size())).integral();
return r;
}
poly arccos() {
poly r = arcsin();
for (int i = 0; i < size(); ++i)
r.a[i] = r.a[i] ? P - r.a[i] : 0;
return r;
}
poly arctan() {
poly sq = (*this) * (*this).modxk(size());
sq.a[0] = 1 + sq.a[0];
if (sq.a[0] >= P)
sq.a[0] -= P;
poly r = (derivative() * sq.inv(size())).integral();
return r;
}
poly arccot() {
poly r = arctan();
for (int i = 0; i < size(); ++i)
r.a[i] = r.a[i] ? P - r.a[i] : 0;
return r;
}
poly mulT(const poly& b) const {
if (b.size() == 0)
return poly();
int n = b.size();
return ((*this) * b.rev()).divxk(n - 1);
}
std::pair<poly, poly> divide(const poly& g) const {
int n = a.size(), m = g.size();
if (n < m)
return make_pair(poly(), a);
poly fR = rev();
poly gR = g.rev().alignxk(n - m + 1);
poly gRI = gR.inv(gR.size());
poly qR = (fR * gRI).modxk(n - m + 1);
poly q = qR.rev();
poly r = ((*this) - g * q).modxk(m - 1);
return std::make_pair(q, r);
}
std::vector<int> eval(std::vector<int> x) const {
if (size() == 0)
return std::vector<int>(x.size(), 0);
const int n = std::max(int(x.size()), size());
std::vector<poly> q(4 * n);
std::vector<int> ans(x.size());
x.resize(n);
std::function<void(int, int, int)> build = [&](int p, int l, int r) {
if (r - l == 1) {
q[p] = std::vector<int>{1, (P - x[l]) % P};
} else {
int m = (l + r) / 2;
build(2 * p, l, m);
build(2 * p + 1, m, r);
q[p] = q[2 * p] * q[2 * p + 1];
}
};
build(1, 0, n);
std::function<void(int, int, int, const poly&)> work = [&](int p, int l, int r,
const poly& num) {
if (r - l == 1) {
if (l < int(ans.size()))
ans[l] = num[0];
} else {
int m = (l + r) / 2;
work(2 * p, l, m, num.mulT(q[2 * p + 1]).modxk(m - l));
work(2 * p + 1, m, r, num.mulT(q[2 * p]).modxk(r - m));
}
};
work(1, 0, n, mulT(q[1].inv(n)));
return ans;
}
};
} // namespace Polynomial
using Polynomial::poly;
void SolveCase(int Case) {
int n, m;
std::cin >> n >> m;
std::vector<int> a(n);
for (int i = 0; i < n; ++i)
std::cin >> a[i];
std::vector<std::pair<poly, poly>> all(n);
std::vector<int> temp(11);
for (int i = 0; i < n; ++i) {
temp[0] = 1;
poly f(temp);
temp[0] = 0;
temp[a[i]] = 1;
poly g(temp);
temp[a[i]] = 0;
all[i] = std::make_pair(f, g);
}
while (all.size() > 1) {
std::vector<std::pair<poly, poly>> next;
for (int i = 0; i < all.size(); i += 2) {
if (i == all.size() - 1) {
next.push_back(all[i]);
} else {
next.push_back({
all[i].first * all[i + 1].first + all[i].second * all[i + 1].second,
all[i].first * all[i + 1].second + all[i].second * all[i + 1].first,
});
}
}
all = std::move(next);
}
poly f = all[0].second;
std::cout << f[m] << "\n";
}
标签：std,AtCoder,return,Beginner,int,poly,267,const,size 来源： https://www.cnblogs.com/zengzk/p/16654142.html
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