AtCoder Beginner Contest 266
咕咕咕咕。
E - Throwing the Die
注意到如果对当前的结果不满意然后重开的话，后面的结果和当前结果是没有关系的。
假设已经算出了有 \(n\) 次机会时的期望得分 \(E_n\)。
那么还剩 \(n + 1\) 次机会的时候，摇了一次结果为 \(X\)，还剩下 \(n\) 次机会。如果 \(E_n > X\)，那么就继续玩；反之则直接结束游戏。
然后 \(E_0 = 0\)，这样子就可以直接递推算了。
AC代码
// Problem: E - Throwing the Die
// Contest: AtCoder - AtCoder Beginner Contest 266
// URL: https://atcoder.jp/contests/abc266/tasks/abc266_e
// 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);
#define freep(p) p ? delete p, p = nullptr, void(1) : void(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;
std::cin >> n;
double e = 0;
for (int i = 1; i <= n; ++i) {
double ne = 0;
for (int j = 1; j <= 6; ++j) {
if (j < e) {
ne = ne + e / 6.0;
} else {
ne = ne + j / 6.0;
}
}
e = ne;
}
std::cout << std::fixed << std::setprecision(12) << e << "\n";
}
F - Well-defined Path Queries on a Namori
询问相当于只能走桥边时两点是否连通，然后tarjan跑出桥边，再并查集维护连通性即可。
AC代码
// Problem: F - Well-defined Path Queries on a Namori
// Contest: AtCoder - AtCoder Beginner Contest 266
// URL: https://atcoder.jp/contests/abc266/tasks/abc266_f
// Memory Limit: 1024 MB
// Time Limit: 3000 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);
#define freep(p) p ? delete p, p = nullptr, void(1) : void(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() {}
std::vector<std::pair<int, int>> EdgeBCC(const std::vector<std::vector<int>>& g) {
int n = g.size();
std::vector<std::pair<int, int>> bridges;
int dfs_clock = 0;
std::vector<int> dfn(n, -1), low(n);
std::function<void(int, int)> tarjan = [&](int u, int fa) {
dfn[u] = low[u] = dfs_clock++;
for (int v : g[u]) {
if (v == fa)
continue;
if (dfn[v] == -1) {
tarjan(v, u);
low[u] = std::min(low[u], low[v]);
if (dfn[u] < low[v]) {
bridges.push_back({u, v});
}
} else if (dfn[v] < dfn[u]) {
low[u] = std::min(low[u], dfn[v]);
}
}
};
for (int i = 0; i < n; ++i) {
if (dfn[i] == -1) {
tarjan(i, i);
}
}
return bridges;
}
void SolveCase(int Case) {
int n;
std::cin >> n;
std::vector<std::vector<int>> g(n);
std::vector<std::pair<int, int>> E;
for (int i = 0; i < n; ++i) {
int u, v;
std::cin >> u >> v;
--u, --v;
g[u].push_back(v);
g[v].push_back(u);
E.push_back({u, v});
}
auto bridges = EdgeBCC(g);
logd(bridges);
std::vector<int> f(n);
std::iota(f.begin(), f.end(), 0);
std::function<int(int)> find = [&](int x) { return x == f[x] ? x : f[x] = find(f[x]); };
for (auto [u, v] : bridges) {
u = find(u), v = find(v);
if (u != v)
f[u] = v;
}
int q;
std::cin >> q;
for (int i = 0; i < q; ++i) {
int u, v;
std::cin >> u >> v;
--u, --v;
u = find(u), v = find(v);
std::cout << (u == v ? "Yes" : "No") << "\n";
}
}
G - Yet Another RGB Sequence
先放 G 和 B ，此时没有任何约束，共 \(\binom{G + B}{G}\) 种方案。
然后，选择 \(K\) 个 G，对于选中的每一个 G ，在前面插一个 R ，共 \(\binom{G}{K}\) 种方案。
此时，有 \(K\) 个 RG 的约束已经满足了，为了满足第一个约束，还需要再插入 \(R - K\) 个 R ，且没有新增的 RG 。这样子的话，一个 R 只能插入到某个 RG 前面，或者某个 B 前面，或者结尾，共 \(K + B + 1\) 个可能的插入位置。
注意：字符是不带标号的，所以不能直接 \({(K + B + 1)}^{R - K}\)，这样子会重复计数。
其实问题转换一下就是，有 \(x\) 个带标号的容器，有 \(y\) 个不带标号的小球，一个容器可以放任意个小球，问有几种不同的放法？这个其实是个挺经典的问题，用隔板法可以得出有 \(\binom{x + y - 1}{x - 1}\) 种放法（用 \(x - 1\) 个隔板把 \(y\) 个小球分成 \(x\) 份）。
AC代码
// Problem: G - Yet Another RGB Sequence
// Contest: AtCoder - AtCoder Beginner Contest 266
// URL: https://atcoder.jp/contests/abc266/tasks/abc266_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);
#define freep(p) p ? delete p, p = nullptr, void(1) : void(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);
}
std::string to_string() const { return std::to_string(value_); }
};
// 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];
}
} binom(3e6 + 5);
void SolveCase(int Case) {
auto fast = [&binom](int r, int g, int b, int k) {
Mint ans = Mint(1);
ans = ans * binom(b + g, b);
// logd(ans);
ans = ans * binom(g, k);
// logd(ans);
ans = ans * binom(k + b + r - k, k + b);
// logd(k + b + r - k, k + b, ans);
return ans.value();
};
int r, g, b, k;
std::cin >> r >> g >> b >> k;
std::cout << fast(r, g, b, k) << "\n";
}
Ex - Snuke Panic (2D)
官方题解写得听清楚的，就座标转换一下变成 1D/1D DP，然后这类DP的转移可以用 CDQ 分治从 \(O(n^2)\) 优化到 \(O(n \log^2 n)\)。
AC代码
// Problem: Ex - Snuke Panic (2D)
// Contest: AtCoder - AtCoder Beginner Contest 266
// URL: https://atcoder.jp/contests/abc266/tasks/abc266_h
// Memory Limit: 1024 MB
// Time Limit: 5000 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);
#define freep(p) p ? delete p, p = nullptr, void(1) : void(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() {}
const i64 INF = INT64_C(0x3f3f3f3f3f3f3f3f);
template <typename T>
class FenwickTree {
public:
using Operator = std::function<T(const T&, const T&)>;
private:
inline int lb(int x) { return x & -x; }
public:
FenwickTree(int n, T init, Operator op = std::plus<T>())
: n_(n), init_(init), c_(n_ + 1, init_), op_(op), t_(n_ + 1, -1), tag_(0) {}
void Update(int x, T d) {
for (; x <= n_; x += lb(x)) {
if (t_[x] != tag_)
c_[x] = init_;
c_[x] = op_(c_[x], d);
t_[x] = tag_;
}
}
T Query(int x) {
T r = init_;
for (; x; x -= lb(x)) {
if (t_[x] == tag_)
r = op_(r, c_[x]);
}
return r;
}
void Reset() { ++tag_; }
T Query(int l, int r) { return Query(r) - Query(l - 1); }
T Kth(int k) {
T ans = 0, cnt = 0;
for (int i = std::__lg(n_) + 1; i >= 0; --i) {
ans += (1LL << i);
if (ans >= n_ || cnt + c_[ans] >= k)
ans -= (1LL << i);
else
cnt += c_[ans];
}
return ans + 1;
}
private:
int n_;
T init_;
std::vector<T> c_;
const Operator op_;
std::vector<int> t_;
int tag_;
};
void SolveCase(int Case) {
int n;
std::cin >> n;
std::vector<std::array<i64, 4>> p;
p.push_back({0, 0, 0, 0});
for (int i = 1; i <= n; ++i) {
i64 t, x, y, a;
std::cin >> t >> x >> y >> a;
p.push_back({y, t - x - y, t + x - y, a});
}
std::vector<i64> dp(n + 1, -INF);
std::function<void(int, int)> CDQ = [&](int l, int r) {
if (l == r)
return;
int mid = (l + r) >> 1;
CDQ(l, mid);
{
std::vector<std::array<i64, 4>> temp;
temp.reserve(r - l + 1);
for (int i = l; i <= r; ++i) {
temp.push_back({p[i][1], p[i][2], i, p[i][3]});
}
std::vector<i64> b, c;
b.reserve(temp.size());
c.reserve(temp.size());
for (auto [x, y, id, a] : temp) {
b.push_back(x);
c.push_back(y);
}
std::sort(b.begin(), b.end());
b.erase(std::unique(b.begin(), b.end()), b.end());
std::sort(c.begin(), c.end());
c.erase(std::unique(c.begin(), c.end()), c.end());
for (auto& [x, y, id, a] : temp) {
x = std::lower_bound(b.begin(), b.end(), x) - b.begin() + 1;
y = std::lower_bound(c.begin(), c.end(), y) - c.begin() + 1;
}
std::sort(temp.begin(), temp.end());
FenwickTree<i64> bit(c.size(), -INF,
[](const i64& x, const i64& y) -> i64 { return std::max(x, y); });
for (auto [x, y, id, a] : temp) {
if (id <= mid) {
bit.Update(y, dp[id]);
} else {
dp[id] = std::max(dp[id], bit.Query(y) + a);
}
}
}
CDQ(mid + 1, r);
};
std::sort(p.begin(), p.end());
dp[0] = 0;
CDQ(0, p.size() - 1);
i64 ans = *std::max_element(dp.begin(), dp.end());
std::cout << ans << "\n";
}
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