比赛总结

发挥基本正常,T3 超级无敌尼玛比大失误

T1 总结

赛时总结

简单题,纯模拟即可

题目思路

模拟

题目代码

T2 总结

赛时总结

一眼题。小学生题目,贪心

题目思路

贪心

题目代码

T3 总结

赛时总结

赛时一眼顶针,结果某一行代码莫名奇妙不见了,导致 RE 还差了半天没查出来

400 -> 325 ,R1 -> R5

题目思路

思路打开,模拟即可

题目代码

T4 总结

赛时总结

一眼顶针二分,场切

题目思路

二分,带点小学奥数(鸡兔同笼)。

题目代码

CSP - S 2023 题解

T4 种树

按照前天的思路成功AC。附上这个首A的代码:

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/**
* @file T4.cpp
* @author Zhoumouren
* @brief
* @version 0.1
* @date 2024-08-15
*
* @copyright Copyright (c) 2024

When We Saw this problem , we can got a easy idea :
Calc the time of grow to a good tree ,
then priority grow the tree what the time longer than others

Then We Found : This Solution is Worng.
We can think about this :
The time grow to Tree is about what the time when the tree grow

So , We can binary search the time of grow to a good tree
then , we can get what time that a tree on land i must grow

this is must a right Solution

*/

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef __int128 int128;

namespace FastOI
{
char buf[1 << 20], *p1, *p2;
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++)
template<typename T> inline T read(T& x) {
x = 0;
int f = 1;
char ch;
while (!isdigit(ch = getchar())) if (ch == '-') f = -1;
while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
x *= f;
return x;
}
template<typename T, typename... Args> inline void read(T& x, Args &...x_) {
read(x);
read(x_...);
return;
}
inline ll read() {
ll x;
read(x);
return x;
}
inline void read(char *s) {
scanf("%s", s + 1);
}
};
using namespace FastOI;

const int N = 1e5 + 10;

struct Edge {
int to, nt;
Edge() {}
Edge(int to, int nt) : to(to), nt(nt) {}
}e[N << 1];
int hd[N], cnte;
void AddEdge(int u, int v) {
e[++cnte] = Edge(v, hd[u]);
hd[u] = cnte;
}

struct Land {
ll a, b, c;
}a[N];
int n;

int fa[N];

int p[N], t[N];
int vis[N];

void Input() {
read(n);
for(int i = 1; i <= n; i++) {
read(a[i].a, a[i].b, a[i].c);
}
int u, v;
for(int i = 1; i < n; i++) {
read(u, v);
AddEdge(u, v);
AddEdge(v, u);
}
}

void Dfs(int u, int f) {
fa[u] = f;
for(int i = hd[u]; i; i = e[i].nt) {
if(e[i].to == f) continue;
Dfs(e[i].to, u);
}
}

int128 Calc(int128 a, int128 b, int128 c) {
ll x = max((int128)1, min(a + 1, (int128)ceil(1.0 * (1 - b) / c)));
if(c < 0) return (x - 1) * b + x * (x - 1) / 2 * c + (a - x + 1);
else if(c == 0) return a * b;
else return (x - 1) + (a - x + 1) * b + (a * (a + 1) / 2 - x * (x - 1) / 2) * c;
}

int update(int u) {
if(vis[u]) return 0;
vis[u] = 1;
if(fa[u]) return update(fa[u]) + 1;
return 1;
}

bool Check(ll x) {
for(int i = 1; i <= n; i++) {
vis[i] = 0;
int128 now = Calc(x, a[i].b, a[i].c);
if (now < a[i].a) return false;
int l = 1, r = n, best = -1;
while(l <= r) {
int mid = l + r >> 1;
if(now - Calc(mid - 1, a[i].b, a[i].c) >= a[i].a) {
l = mid + 1;
best = mid;
}
else {
r = mid - 1;
}
}
p[i] = i, t[i] = best, vis[i] = 0;
}
sort(p + 1, p + n + 1, [](int A, int B) {return t[A] < t[B];});
int cnt = 0;
for(int i = 1; i <= n; i++) {
cnt += update(p[i]);
if(cnt > t[p[i]]) return false;
}
return true;
}

void Work() {
Dfs(1, 0);
// vis[0] = 1;
ll l = n, r = 1e9, best = -1;
while(l <= r) {
ll mid = l + r >> 1;
if(Check(mid)) {
r = mid - 1;
best = mid;
}
else {
l = mid + 1;
}
}
printf("%lld", best);
}

int main() {
int T = 1;
while(T--) {
Input();
Work();
}
return 0;
}