# Codeforces Round #591

## Contest Info

Solved A B C D E F
4/6 O O O O - -
• O 在比赛中通过
• Ø 赛后通过
• ! 尝试了但是失败了
• - 没有尝试

## Solutions

### A. Save the Nature

$n$个数$p_i$

• 如果一个下标它是$a$的倍数，那么这个位置的价值就是$a_i \cdot x\%$
• 如果一个下标它是$b$的倍数，那么这个位置的价值就是$a_i \cdot y\%$
• 如果一个下标它是$lcm(a, b)$的倍数，那么这个位置的价值就是$a_i \cdot (x + y)\%$

view code
#include <bits/stdc++.h>
#define debug(...) { printf("#  "); printf(__VA_ARGS__); puts(""); }
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << " "; pt(args...); }
template <class T> inline void pt(const T &s) { cout << s << "\n"; }
template <class T> inline void pt(const vector <T> &vec) { for (auto &it : vec) cout << it << " "; cout << endl; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
constexpr int N = 2e5 + 10;
int n, p[N], x, y, a, b; ll k;
bool check(ll mid) {
ll lcm = 1ll * a * b / gcd(a, b);
ll cnt[3] = {mid / a - mid / lcm, mid / b - mid / lcm, mid / lcm};
ll tot = 0;
for (int i = 1; i <= mid; ++i) {
if (cnt[2]) {
tot += 1ll * p[i] / 100 * (x + y);
--cnt[2];
} else if (cnt[1]) {
tot += 1ll * p[i] / 100 * y;
--cnt[1];
} else if (cnt[0]) {
tot += 1ll * p[i] / 100 * x;
--cnt[0];
} else break;
}
}
void run() {
cin >> n;
for (int i = 1; i <= n; ++i) cin >> p[i];
sort(p + 1, p + 1 + n, [&](int x, int y) { return x > y; });
cin >> x >> a >> y >> b >> k;
if (x > y) swap(x, y), swap(a, b);
int l = 1, r = n, res = -1;
while (r - l >= 0) {
int mid = (l + r) >> 1;
if (check(mid)) {
res = mid;
r = mid - 1;
} else
l = mid + 1;
}
pt(res);
}

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T; cin >> _T;
while (_T--) run();
return 0;
}


### B. Sequence Sorting

view code
#include <bits/stdc++.h>
#define debug(...) { printf("#  "); printf(__VA_ARGS__); puts(""); }
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << " "; pt(args...); }
template <class T> inline void pt(const T &s) { cout << s << "\n"; }
template <class T> inline void pt(const vector <T> &vec) { for (auto &it : vec) cout << it << " "; cout << endl; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
constexpr int N = 3e5 + 10;
int n, a[N], l[N], r[N];
void run() {
cin >> n;
memset(l, 0x3f, sizeof (l[0]) * (n + 10));
memset(r, 0, sizeof (r[0]) * (n + 10));
for (int i = 1; i <= n; ++i) {
cin >> a[i];
chmin(l[a[i]], i);
chmax(r[a[i]], i);
}
int res = 1, tot = 0;
int lst = 0, sum = 0;
for (int i = 1; i <= n; ++i) if (r[i] != 0) {
++tot;
if (l[i] > lst) {
lst = r[i];
++sum;
chmax(res, sum);
} else {
lst = r[i];
sum = 1;
}
}
pt(tot - res);
}

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T; cin >> _T;
while (_T--) run();
return 0;
}


### C. Paint the Tree

$f[i][0]$表示$i$的子树中并且$i$没有一种颜色可以涂的最大价值，$f[i][1]$表示$i$的子树中并且$i$还剩了至少一种颜色可以涂的最大价值。

view code
#include <bits/stdc++.h>
#define debug(...) { printf("#  "); printf(__VA_ARGS__); puts(""); }
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << " "; pt(args...); }
template <class T> inline void pt(const T &s) { cout << s << "\n"; }
template <class T> inline void pt(const vector <T> &vec) { for (auto &it : vec) cout << it << " "; cout << endl; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
constexpr int N = 5e5 + 10;
int n, k, fa[N]; ll f[N][2], g[N];
vector <vector<pII>> G;
void dfs(int u) {
f[u][0] = f[u][1] = 0;
for (auto &it : G[u]) {
int v = it.fi, w = it.se;
if (v == fa[u]) continue;
fa[v] = u;
dfs(v);
g[v] = f[v][1] + w - f[v][0];
f[u][0] += f[v][0];
f[u][1] += f[v][0];
}
vector <int> id;
for (auto &it : G[u]) if (it.fi != fa[u]) id.push_back(it.fi);
sort(id.begin(), id.end(), [&](int x, int y) { return g[x] > g[y]; });
for (int i = 0, sze = id.size(); i < k && i < sze; ++i) {
int v = id[i];
if (g[v] < 0) break;
if (i < k - 1) f[u][1] += g[v];
f[u][0] += g[v];
}
}
void run() {
cin >> n >> k;
G.clear(); G.resize(n + 1);
for (int i = 1, u, v, w; i < n; ++i) {
cin >> u >> v >> w;
G[u].emplace_back(v, w);
G[v].emplace_back(u, w);
}
fa[1] = 1;
dfs(1);
pt(max(f[1][0], f[1][1]));
}

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T; cin >> _T;
while (_T--) run();
return 0;
}


### D. Stack Exterminable Arrays

• 如果栈空或者栈顶不等于当前数$a_i$，那么将$a_i$加入到栈顶
• 如果栈顶的数等于当前数$a_i$，那么栈顶弹出。

$S[i]$哈希一下即可。

view code
#include <bits/stdc++.h>
#define debug(...) { printf("#  "); printf(__VA_ARGS__); puts(""); }
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << " "; pt(args...); }
template <class T> inline void pt(const T &s) { cout << s << "\n"; }
template <class T> inline void pt(const vector <T> &vec) { for (auto &it : vec) cout << it << " "; cout << endl; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
constexpr int N = 3e5 + 10;
//
struct Hash {
static ull base[N];
static void init() {
base[0] = 1;
for (int i = 1; i < N; ++i)
base[i] = base[i - 1] * 19260817;
}
ull a[N];
inline void work() { a[0] = 0; }
inline void add(int ch, int id) {
a[id] = a[id - 1] * 19260817 + ch;
}
}hs;
ull Hash::base[N] = {0};
map <ull, int> mp;
int n, a[N], sta[N];
void run() {
cin >> n;
for (int i = 1; i <= n; ++i) cin >> a[i];
mp.clear();
mp[0] = 1; *sta = 0;
ll res = 0;
hs.work();
for (int i = 1; i <= n; ++i) {
if (*sta && sta[*sta] == a[i]) --*sta;
else sta[++*sta] = a[i], hs.add(a[i], *sta);
ull H = hs.a[*sta]; res += mp[H];
++mp[H];
}
pt(res);
}

int main() {
Hash::init();
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T; cin >> _T;
while (_T--) run();
return 0;
}

posted @ 2019-10-10 09:33  Dup4  阅读(367)  评论(0编辑  收藏