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Shiroha白羽的博客

Golang 踩坑 —— interface 为参数的时候传 nil 指针 Codeforces Round 925 (Div. 3) Codeforces Round 924 (Div. 2) Codeforces Round 923 (Div. 3) Codeforces Round 922 (Div. 2) Codeforces Round 921 (Div. 2) Educational Codeforces Round 161 (Rated for Div. 2) Codeforces Round 920 (Div. 3) Codeforces Round 919 (Div. 2) Hello 2024 Good Bye 2023 Codeforces Round 918 (Div. 4) 个人备份的常用 macOS 清理命令 Codeforces Round 917 (Div. 2) Pinely Round 3 (Div. 1 + Div. 2) Educational Codeforces Round 160 (Rated for Div. 2) Codeforces Round 915 (Div. 2) Codeforces Round 914 (Div. 2) Codeforces Round 913 (Div. 3) Educational Codeforces Round 159 (Rated for Div. 2) Codeforces Round 912 (Div. 2) Codeforces Round 911 (Div. 2) CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!) Educational Codeforces Round 158 (Rated for Div. 2) Codeforces Round 910 (Div. 2) Codeforces Round 909 (Div. 3) Codeforces Round 908 (Div. 2) Educational Codeforces Round 157 (Rated for Div. 2) C++自定义的字面量 Codeforces Round 907 (Div. 2) Codeforces Round 916 (Div. 3) 关于 LRU map 的一些灵感 2023 杭州站 ICPC 现场赛 反复横跳的 Clang-Tidy(cert-dcl21-cpp) Codeforces Round 906 (Div. 2) 一段奇怪的 CPP 代码 Codeforces Round 905 (Div. 3) Codeforces Round 904 (Div. 2) Codeforces Round 903 (Div. 3) Educational Codeforces Round 156 (Rated for Div. 2) Codeforces Round 902 (Div. 2, based on COMPFEST 15 - Final Round) Codeforces Round 901 (Div. 2) Codeforces Round 900 (Div. 3) Codeforces Round 899 (Div. 2) Educational Codeforces Round#155 (Div. 2) Codeforces Round 898 (Div. 4) CodeTON Round 6 (Div. 2) Codeforces Round 897 (Div. 2) Codeforces Round 896 (Div. 2) Codeforces Round 887 (Div. 2) Codeforces Round 895 (Div. 3) 左值-右值-将亡值 blog.mauve.icu Pinely Round 2 (Div. 1 + Div. 2) Harbour.Space Scholarship Contest 2023-2024 (Div. 1 + Div. 2) Codeforces Round 894 (Div. 3) Codeforces Round 888 (Div. 3) Educational Codeforces Round#153 (Div. 2) Codeforces Round 893 (Div. 2) OTPAUTH,两步验证中的通用协议 Codeforces Round 892 (Div. 2) Codeforces Round 891 (Div. 3) Codeforces Round 890 (Div. 2) Educational Codeforces Round#152 (Div. 2) blog.mauve.icu Java Script 的 null 和 undefined 随想 记一次 SQL LEFT JOIN 没有得到预期结果的错误 Codeforces Round#789(Div. 2) GCC/G++ 预编译头性能优化 使用 Junit5 和 Mockito 实现 SpringBoot 的单元测试最优美的解决方案 centOS 防火墙 docker-compse 的问题 C++ 语言实现动态变化的线程池 Codeforces Round#744 (Div. 3) 计算机图形学 Windows 通过网络访问 WSL2 原生 JavaScript 实现图片裁剪 面试复习(计算机图形学) 面试复习(算法) Codeforces Round#706(Div. 2)-Let's Go Hiking 面试复习(Java) 面试复习(Git) 面试复习(Linux) 面试复习(数据库) 面试复习(计算机网络) 面试复习(操作系统) 面试复习(C++) Codeforces Round#699 (Div. 2) 清理 WSL2 的磁盘占用 Codeforces Round#697 (Div. 3) Windows 下的 NTFS 驱动器索引 BUG 计算机网络复习 记一次 Navicat 连接 MySQL 一直报认证错误(Access denied) 计算机网络实验复习 WSL1 使用 Docker 一直无法启动 2020牛客暑期多校训练营(第三场)D-Points Construction Problem——构造 2020牛客暑期多校训练营(第三场)E-Two Matchings——复杂思维与简单dp 2020牛客暑期多校训练营(第二场)I-Interval——最大流转对偶图求最短路 Educational Codeforces Round 80 D. Minimax Problem——二分+二进制处理 Codeforces Round 606 E. Two Fairs——图论 Codeforces Round 612 (Div. 2) C. Garland——DP
Codeforces Round 926 (Div. 2)
Shiroha · 2024-04-13 · via Shiroha白羽的博客

A. Sasha and the Beautiful Array

大致题意

有一个数组,现在允许你任意排序它,使得其所有的相邻对之差之和最小,问如何操作

思路

排序一下就行,这样就等于最大的那个值减去最小的那个

AC code

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void solve() {
int _;
cin >> _;
for (int tc = 0; tc < _; ++tc) {
int n;
cin >> n;
vector<int> data(n);
for (auto &i: data) cin >> i;
sort(data.begin(), data.end());
cout << data.back() - data.front() << endl;
}
}

B. Sasha and the Drawing

大致题意

有一个正方形,其上有 $4 \times n - 2$ 条对角线,现在要你染黑一些格子,使得这些对角线至少有 $x$ 个被覆盖,问最少染黑几个

思路

只要染黑第一行和最下面一行即可,显然,除了四个角落,其他几个点染了就是影响两条对角线

AC code

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void solve() {
int _;
cin >> _;
for (int tc = 0; tc < _; ++tc) {
int n, m;
cin >> n >> m;
if (m <= 4 * n - 4) cout << (m + 1) / 2 << endl;
else if (m == 4 * n - 3) cout << 2 * n - 1 << endl;
else if (m == 4 * n - 2) cout << 2 * n << endl;
}
}

C. Sasha and the Casino

大致题意

在赌场赌博,已知每次可以下注任意合理的钱 $y$,赢了就收回 $k \times y$,输了就没了,且最多连续输 $x$ 场,问是否赚到任意数量的钱

思路

根据赌徒原理做,要保证你每次下注的时候,如果赢了能把之前输的钱全都赚回来,且还要多赚一点

AC code

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#define int long long

void solve() {
int _;
cin >> _;
for (int tc = 0; tc < _; ++tc) {
int k, x, a;
cin >> k >> x >> a;
int ca = a, lose = 0;
bool flag = true;
for (int i = 0; i < x; ++i) {
int cur = (lose + k - 1) / (k - 1);
if (ca < cur) {
flag = false;
break;
}
ca -= cur;
lose += cur;
}
if (ca * k <= a) flag = false;
cout << (flag ? "YES" : "NO") << endl;
}
}

D. Sasha and a Walk in the City

大致题意

有一棵树,现在要选择一定数量的节点染色,使得任意两个节点之间的路径最多只经过两个染黑节点,问如何操作

思路

树上 dp 即可,关注当前节点到所有下面的子节点中,染色数量最多的路径染色了多少个,可以枚举 1 个和 2 个的情况(0 个一定只有一种)

AC code

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#define int long long

void solve() {
int _;
cin >> _;
for (int tc = 0; tc < _; ++tc) {
int n;
cin >> n;
constexpr int mod = 998244353;
vector<pair<int, int>> edges((n - 1) * 2);
vector<int> head(n + 1, -1);
for (int i = 0; i < n - 1; ++i) {
int u, v;
cin >> u >> v;
edges[i << 1] = {v, head[u]};
edges[i << 1 | 1] = {u, head[v]};
head[u] = i << 1;
head[v] = i << 1 | 1;
}

function<pair<int, int>(int, int)> dfs = [&](int u, int p) {
int a = 1, b = 0;
for (int e = head[u]; ~e; e = edges[e].second) {
if (edges[e].first == p) continue;
auto [na, nb] = dfs(edges[e].first, u);
a = (a * (1 + na)) % mod;
b = (b + na + nb) % mod;
}

return make_pair(a, b);
};

auto [a, b] = dfs(1, 0);
cout << (a + b + 1) % mod << endl;
}
}