Codeforces Round 998 (Div. 3)
A. Fibonacciness
题意:给你
枚举
点击查看代码
void solve() {
int a, b, c, d;
std::cin >> a >> b >> c >> d;
int ans1 = 1;
if (b + a + b == c) {
++ ans1;
}
if (a + b + c == d) {
++ ans1;
}
int ans2 = 1;
if (b + d - c == c) {
++ ans2;
}
if (a + b == d - c) {
++ ans2;
}
int ans3 = 1;
if (a + b == c - b) {
++ ans3;
}
if (c - b + c == d) {
++ ans3;
}
std::cout << std::max({ans1, ans2, ans3}) << "\n";
}
B. Farmer John's Card Game
题意:
要出完所有牌,那么打出的牌肯定是
点击查看代码
void solve() {
int n, m;
std::cin >> n >> m;
std::vector<int> a(n * m);
for (int i = 0; i < n; ++ i) {
for (int j = 0; j < m; ++ j) {
int x;
std::cin >> x;
a[x] = i;
}
}
std::vector<int> ans(n), last(n, -n);
for (int i = 0; i < n * m; ++ i) {
if (i - last[a[i]] < n) {
std::cout << -1 << "\n";
return;
}
last[a[i]] = i;
if (i < n) {
ans[i] = a[i];
}
}
for (int i = 0; i < n; ++ i) {
std::cout << ans[i] + 1 << " \n"[i == n - 1];
}
}
C. Game of Mathletes
有n个数,每次
分数是固定的,因为如果牌库里有
点击查看代码
void solve() {
int n, k;
std::cin >> n >> k;
std::vector<int> cnt(2 * n + 1);
for (int i = 0; i < n; ++ i) {
int x;
std::cin >> x;
++ cnt[x];
}
int ans = 0;
for (int i = 1; i <= k / 2; ++ i) {
if (k % 2 == 0 && i == k / 2) {
ans += cnt[i] / 2;
} else {
ans += std::min(cnt[i], cnt[k - i]);
}
}
std::cout << ans << "\n";
}
D. Subtract Min Sort
题意:一个数组
假设
注意特判
点击查看代码
void solve() {
int n;
std::cin >> n;
std::vector<int> a(n);
for (int i = 0; i < n; ++ i) {
std::cin >> a[i];
}
if (a[0] > a[1]) {
std::cout << "NO\n";
return;
}
for (int i = 1; i + 1 < n; ++ i) {
a[i] -= a[i - 1];
if (a[i] > a[i + 1]) {
std::cout << "NO\n";
return;
}
}
std::cout << "YES\n";
}
E. Graph Composition
题意:给你两个图
我们给两个图都开一个并查集记录集合。我们用
首先如果
然后枚举
点击查看代码
void solve() {
int n, m, k;
std::cin >> n >> m >> k;
std::set<std::pair<int, int> > s1, s2;
std::vector<int> fa1(n + 1), fa2(n + 1);
std::iota(fa1.begin(), fa1.end(), 0);
std::iota(fa2.begin(), fa2.end(), 0);
std::function<int(int)> find1 = [&](int x) -> int {
return x == fa1[x] ? x : fa1[x] = find1(fa1[x]);
};
std::function<int(int)> find2 = [&](int x) -> int {
return x == fa2[x] ? x : fa2[x] = find2(fa2[x]);
};
for (int i = 0; i < m; ++ i) {
int u, v;
std::cin >> u >> v;
s1.insert({u, v});
}
for (int i = 0; i < k; ++ i) {
int u, v;
std::cin >> u >> v;
fa2[find2(u)] = find2(v);
s2.insert({u, v});
}
int ans = 0;
for (auto & [x, y] : s1) {
if (find2(x) != find2(y)) {
++ ans;
} else {
fa1[find1(x)] = find1(y);
}
}
std::map<int, std::vector<int> > mp;
for (int i = 1; i <= n; ++ i) {
mp[find2(i)].push_back(i);
}
for (int i = 1; i <= n; ++ i) {
for (auto & x : mp[i]) {
if (find1(x) != find1(i)) {
++ ans;
fa1[find1(x)] = find1(i);
}
}
}
std::cout << ans << "\n";
}
F. Multiplicative Arrays
题意:求有多少数组
注意到因为
那么
有了
点击查看代码
void solve() {
int n, k;
std::cin >> k >> n;
std::vector<std::vector<int> > factor(k + 1);
for (int i = 2; i <= k; ++ i) {
for (int j = i; j <= k; j += i) {
factor[j].push_back(i);
}
}
int m = std::__lg(k) + 1;
std::vector f(k + 1, std::vector<Z>(m + 1));
for (int i = 1; i <= k; ++ i) {
for (int j = 1; j <= m; ++ j) {
if (i > 1 && j == 1) {
f[i][j] = 1;
} else {
for (auto & x : factor[i]) {
f[i][j] += f[i / x][j - 1];
}
}
}
}
std::vector<Z> ans(k + 1);
ans[1] = n;
for (int i = 2; i <= k; ++ i) {
for (int j = 1; j <= std::min(m, n); ++ j) {
Z C = 1;
for (int a = 1, b = n + 1; a <= j + 1; ++ a, -- b) {
C = C * b / a;
}
ans[i] += C * f[i][j];
}
}
for (int i = 1; i <= k; ++ i) {
std::cout << ans[i] << " \n"[i == k];
}
}
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】博客园社区专享云产品让利特惠,阿里云新客6.5折上折
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 在鹅厂做java开发是什么体验
· 百万级群聊的设计实践
· WPF到Web的无缝过渡:英雄联盟客户端的OpenSilver迁移实战
· 永远不要相信用户的输入:从 SQL 注入攻防看输入验证的重要性
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析