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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 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;
}
}