模拟赛SXJ202507270900比赛记录&题解
题目请看T1
贪心：主要考察\(50\%\)时\(差值\ mod \ 3 \neq 0\)的情况
\(\begin{cases}\text{计算 } cha = 50 - n \\\text{如果 } cha \bmod 2 \neq 0 \text{ 则} \\\quad \text{输出 } \left\lceil \dfrac{cha}{2} \right\rceil + 1 \\\text{否则} \\\quad \text{输出 } \dfrac{cha}{2} \\\end{cases}\)
\(\begin{cases}\quad \text{计算 } cha = n - 50 \\\quad \begin{cases}\text{如果 } cha \bmod 3 = 1 \text{ 则} & \text{输出 } \dfrac{cha - 1}{3} + 2 \\\text{如果 } cha \bmod 3 = 2 \text{ 则} & \text{输出 } \dfrac{cha - 2}{3} + 4 \\\text{如果 } cha \bmod 3 = 0 \text{ 则} & \text{输出 } \dfrac{cha}{3} \\\end{cases} \\\end{cases}\)
贴一下代码#include <bits/stdc++.h>
using namespace std;
#define fre(c) freopen(c".in","r",stdin);freopen(c".out","w",stdout);
#define ll long long
#define endl "\n"
#define ios ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define cst const
int t, jia = 2, jian = 3;
ll read() {
char c;
ll sum = 0;
int f = 1;
c = getchar();
while (!isdigit(c)) {
if (c == '-')
f *= -1;
c = getchar();
}
while (isdigit(c)) {
sum = sum * 10 + (c - '0');
c = getchar();
}
return sum * f;
}
void fun() {
int n;
cin >> n;
if (n == 50) {
cout << "0" << endl;
}
if (n < 50) {
int cha = 50 - n;
if (cha % jia != 0) {
cha += jian;
cout << cha / 2 + 1 << endl;
} else {
cout << cha / 2 << endl;
}
}
if (n > 50) {
int cha = n - 50;
if (cha % jian == 1) {
cout << (cha - 1) / jian + 2 << endl;
}
if (cha % jian == 2) {
cout << (cha - 2) / jian + 4 << endl;
}
if (cha % jian == 0) {
cout << cha / jian << endl;
}
}
}
int main() {
ios
// fre("3")
fre("light")
cin >> t;
while (t --) {
fun();
}
return 0;
}可惜：TLE 40point
正确做法就是每次把充能值在根节点相加，那么根节点的叶子节点都会有\(父亲节点+自己单独的充能\)
贴一下代码#include using namespace std;#define fre(c) freopen(c".in","r",stdin);freopen(c".out","w",stdout);#define ll long long#define endl "\n"#define ios ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);#define cst constcst int N = 2 * 1e5 + 5;int n, q, head, tot = 0, fa;ll val;struct Edge { int to, nxt;} tree;void link(int x, int y) { tot ++; tree.to = y; tree.nxt = head; head = tot;}void dfs(int u, int root) { for (int i = head; i; i = tree.nxt) { int v = tree.to; if (v == root) { continue ; } val += val; dfs(v, u); }}int main() { ios fre("tree") cin >> n >> q; for (int i = 1; i > u >> v; link(u, v); link(v, u); fa = u; } while (q --) { int p; ll x; cin >> p >> x; val += x;// dfs(p, fa, x); } dfs(1, -1); for (int i = 1; i
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