output stringlengths 52 181k | instruction stringlengths 296 182k |
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#include <bits/stdc++.h>
using namespace std;
long long ans;
int dp[300000 + 5][20], n;
vector<int> e[300000 + 5];
inline void upd(int &x, int y) {
if (x < y) x = y;
}
int dfs(int now, int fa) {
int mxdep = 0;
for (auto v : e[now])
if (v != fa) upd(mxdep, dfs(v, now));
mxdep++;
ans += mxdep;
dp[now][1] ... | ### Prompt
Your task is to create a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
vector<int> e[N];
int n, v, f[N], g[N], q[N];
long long sum;
void dfs(int x, int fa, int v) {
f[x] = 1;
g[x] = 1;
for (auto i : e[x])
if (i != fa) {
dfs(i, x, v);
g[x] = max(g[x], g[i]);
}
if (e[x].size() - (fa != 0) >= v) {... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
const long long N = 300005;
long long n, dp[N][20][2], ans, h[N];
vector<vector<long long> > gr;
priority_queue<long long> pq[N][20];
void dfs(long long u, long long par) {
h[u] = 1;
dp[u][1][0] = dp[u][1][1] = n;
for (long long i = 0; i < gr[u].size(); ++i) {
lon... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
long long int n;
vector<vector<int> > child(300001, vector<int>());
bool was[300001];
int DP3till[300001];
int DP2till[300001];
int k2[300001];
long long int toadd;
pair<int, int> dfs(int v, int k) {
multiset<int> ways;
int cn1 = 0;
int mx = 0;
for (int u : child[v]... | ### Prompt
Generate a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
long long n, fh[300069], dp[300069][18], z;
vector<long long> al[300069];
bitset<300069> vtd;
void bd(long long x) {
long long i, j, sz = al[x].size(), l, lh, rh, md, zz, c;
vector<long long> v;
vtd[x] = 1;
for (i = 0; i < sz; i++) {
l = al[x][i];
if (!vtd[l... | ### Prompt
Construct a Cpp code solution to the problem outlined:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of i... |
#include <bits/stdc++.h>
using namespace std;
int const M = 3e5 + 100, inf = 1e9 + 20, mod = 1e9 + 7;
int n, a[M], dp[M], st[M], fin[M], mark[M];
vector<pair<int, int> > cand;
vector<int> ve;
vector<int> adj[M];
vector<pair<int, int> > good;
int par[M][30];
long long ans;
vector<int> hist;
bool seg[M * 4];
int pw(int x... | ### Prompt
In CPP, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
int read() {
int x = 0;
bool flg = false;
char ch = getchar();
for (; !isdigit(ch); ch = getchar())
if (ch == '-') flg = true;
for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
return flg ? -x : x;
}
int n;
long long ans;
struct Edge {... | ### Prompt
Create a solution in Cpp for the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ... |
#include <bits/stdc++.h>
using namespace std;
inline char gc() {
static char buf[100000], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
? EOF
: *p1++;
}
inline int read() {
int x = 0;
char ch = getchar();
bool positive = 1;
for ... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 10;
const int MAX_LOG = 20;
vector<int> vertex[MAXN];
vector<int> e;
int h[MAXN];
int z[MAXN];
int dp[MAXN];
bool check[MAXN];
long long res;
void dfs(int v) {
check[v] = true;
h[v] = 1;
e.clear();
for (int i = 0; i < vertex[v].size(); i++) {
... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffff;
const long long SLINF = 0x7fffffffffffffff;
const int MAXN = 300007;
int n;
long long ans;
vector<int> ch[MAXN];
vector<int> dp[MAXN];
vector<int> tmp[MAXN];
void init();
voi... | ### Prompt
Please create a solution in Cpp to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
inline int read() {
int x;
char c;
while ((c = getchar()) < '0' || c > '9')
;
for (x = c - '0'; (c = getchar()) >= '0' && c <= '9';) x = x * 10 + c - '0';
return x;
}
struct edge {
int nx, t;
} e[300000 * 2 + 5];
int n, k, h[300000 + 5], en, c[300000 + 5], c... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
inline void upd1(T1& a, T2 b) {
a > b ? a = b : 0;
}
template <class T1, class T2>
inline void upd2(T1& a, T2 b) {
a < b ? a = b : 0;
}
struct ano {
operator long long() {
long long x = 0, y = 0, c = getchar();
while (c < 48) y = ... | ### Prompt
Please create a solution in Cpp to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
int n, sl, fh, cnt, s[1000010], fa[1000010], dp[1000010], mxd[1000010],
f[1000010][20];
long long res, ans;
int t, h[1000010];
struct Tre {
int to, nxt;
} e[1000010 << 1];
vector<pair<int, int> > vt[1000010];
int rd() {
sl = 0;
fh = 1;
char ch = getchar();
whi... | ### Prompt
Please create a solution in CPP to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
int N;
vector<int> G[300005];
int PosRes[22][300005], Pos[22][300005];
int V[22][300005];
int DP[300005];
vector<int> Who;
long long ans;
int Cnt[30];
void Read() {
scanf("%d", &N);
for (int i = 1; i < N; i++) {
int x, y;
scanf("%d%d", &x, &y);
G[x].push_bac... | ### Prompt
Create a solution in Cpp for the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ... |
#include <bits/stdc++.h>
using namespace std;
int n, head[300005], num, x, y, dep[300005], Max[300005][21], own[300005][21];
long long ans;
struct edge {
int to, nxt;
} a[300005 << 1];
void add(int x, int y) { a[++num] = (edge){y, head[x]}, head[x] = num; }
bool check(int x, int f, int id, int y) {
int res = 0;
f... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
x = 0;
char c = getchar(), f = 0;
for (; c < 48 || c > 57; c = getchar())
if (!(c ^ 45)) f = 1;
for (; c >= 48 && c <= 57; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
f ? x = -x : x;
}
const int N = 300005;... | ### Prompt
Your task is to create a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void in(T &x) {
T c = getchar();
while (((c < 48) || (c > 57)) && (c != '-')) c = getchar();
bool neg = false;
if (c == '-') neg = true;
x = 0;
for (; c < 48 || c > 57; c = getchar())
;
for (; c > 47 && c < 58; c = getchar()) x = (x *... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * inf;
const int N = 300000 + 7;
const int M = 20;
const int multipleTest = 0;
vector<int> adj[N];
int n;
long long ans = 0;
int maxLen[N];
int maxK[N][M];
int maxP[N][M];
bool check(int u, int len, int k, int r) {
... | ### Prompt
Construct a Cpp code solution to the problem outlined:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of i... |
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, M = 20;
long long ans, sum;
int n, h[N], edge[N << 1], nxt[N << 1], cnt, f[N][M], g[N], d[N], dep[N], p, q,
fa[N], a[N];
bool vis[N];
vector<pair<int, int> > hep[N];
inline void add(int x, int y) {
edge[++cnt] = y;
nxt[cnt] = h[x];
h[x] = cn... | ### Prompt
Please formulate a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5;
int n, dp[N][19];
long long ans;
vector<int> g[N];
int intial(int u, int p) {
int ret = 0;
for (auto i : g[u])
if (i != p) ret = max(ret, intial(i, u));
ans += ret + 1;
return ret + 1;
}
void dfs(int u, int p) {
bool leaf = 1;
dp[u][1] = n... | ### Prompt
Your task is to create a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
const long long N = 3e5 + 2;
const long long inf = 1e9 + 7;
long long dp[N][22], f[N], max1[N][22];
long long ans = 0;
vector<long long> adj[N];
void dfs(long long x, long long p) {
vector<long long> temp[22];
long long i, j, lef, rig, mid;
for (i = 0; i < adj[x].size... | ### Prompt
In cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
const int NMAX = 3e5 + 10, LGMAX = 19;
int N;
int MaxLevel;
int64_t answer;
int DP[LGMAX][NMAX], father[NMAX], val[NMAX];
vector<int> T[NMAX], LevelNodes[NMAX];
vector<pair<int, int>> GoUp[NMAX];
int DFS(int node, int from, int level) {
father[node] = from;
MaxLevel = m... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
bool cmax(T& a, const U& b) {
return a < b ? a = b, 1 : 0;
}
template <class T, class U>
bool cmin(T& a, const U& b) {
return b < a ? a = b, 1 : 0;
}
void _BG(const char* s) { cerr << s << endl; };
template <class T, class... TT>
void _BG(con... | ### Prompt
In CPP, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
return (b == 0LL ? a : gcd(b, a % b));
}
long double dist(long double x, long double arayikhalatyan, long double x2,
long double y2) {
return sqrt((x - x2) * (x - x2) +
(arayikhalatyan ... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
const int N = 300000 + 10, M = 20 + 2;
int n;
vector<int> V[N];
int depth[N];
int d[M][N], d_max[M][N];
void dfs(int v, int parent) {
depth[v] = 1;
for (int i = 0; i < (((int)((V[v]).size()))); ++i) {
int to = V[v][i];
... | ### Prompt
Develop a solution in Cpp to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
long long ans;
int dp[300000 + 5][20], n;
vector<int> e[300000 + 5];
void upd(int &x, int y) {
if (x < y) x = y;
}
int dfs(int now, int fa) {
int mxdep = 0;
for (auto v : e[now])
if (v != fa) upd(mxdep, dfs(v, now));
mxdep++;
ans += mxdep;
dp[now][1] = n;
... | ### Prompt
Please provide a cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
int n;
const int MaxN = 3e5;
vector<int> adj[MaxN];
vector<int> prec[MaxN];
void dfs(const int u, const int p) {
prec[u] = {1, n};
adj[u].erase(remove(adj[u].begin(), adj[u].end(), p), adj[u].end());
for (auto &&v : adj[u]) {
dfs(v, u);
prec[u][0] = max(prec[u... | ### Prompt
Your task is to create a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
int head[524288], last[1048576], to[1048576], cnt = 0;
void add(int u, int v) {
cnt++;
last[cnt] = head[u];
head[u] = cnt;
to[cnt] = v;
}
int dp[524288];
int a[524288];
long long answer = 0;
int n;
void dfs1(int u, int f, int k) {
for (int i = head[u]; i; i = last... | ### Prompt
Please formulate a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * inf;
const int N = 300000 + 7;
const int M = 19;
const int multipleTest = 0;
vector<int> adj[N];
int n;
long long ans = 0;
int maxLen[N];
int maxK[N][M];
int maxP[N][M];
bool check(int u, int len, int k, int r) {
... | ### Prompt
Please create a solution in CPP to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
inline char gc() {
static char buf[100000], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
? EOF
: *p1++;
}
inline int read() {
int x = 0;
char ch = getchar();
bool positive = 1;
for ... | ### Prompt
Please create a solution in cpp to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3E5 + 77;
const int log2maxn = 23;
struct EDGE {
int to, next;
} edges[maxn * 2];
int cEdge = 1, head[maxn];
void addEdge(int from, int to) {
edges[cEdge] = (EDGE){to, head[from]};
head[from] = cEdge++;
}
int N, V[maxn][log2maxn];
int X[maxn], Y[maxn]... | ### Prompt
Please create a solution in Cpp to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
int m, k;
long long ans;
const int N = 330000;
int h[N], dp[N], mdp[N], cnt[N], par[N], chd[N];
vector<int> adj[N];
vector<int> V;
void dfs(int u, int p) {
V.push_back(u);
par[u] = p;
for (int v : adj[u]) {
if (v == p) continue;
dfs(v, u);
}
chd[u] = adj[u... | ### Prompt
Please create a solution in CPP to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int mod1 = 1e9 + 7, mod2 = 998244353, maxn = 3e5 + 5, maxlog = 20, K = 26;
const long long infll = 1e18;
const double pi = acos(-1);
int n, h[maxn], f[maxn][20], g[maxn][20];
long long res = 0;
vector<int> gr[maxn];
void dfs(int u, int pa) {
for (auto v : gr[u]) {
... | ### Prompt
Your task is to create a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 300005;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b... | ### Prompt
Create a solution in Cpp for the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ... |
#include <bits/stdc++.h>
using namespace std;
const long long int Maxn3 = 1e3 + 10;
const long long int Maxn4 = 1e4 + 10;
const long long int Maxn5 = 1e5 + 10;
const long long int Maxn6 = 1e6 + 10;
const long long int Maxn7 = 1e7 + 10;
const long long int Maxn8 = 1e8 + 10;
const long long int Maxn9 = 1e9 + 10;
const lo... | ### Prompt
In Cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void maxify(T& x, const T& y) {
y > x && (x = y);
}
typedef const int& ci;
struct node {
int v, nxt;
__inline__ __attribute__((always_inline)) node() {}
__inline__ __attribute__((always_inline)) node(ci _v, ci _nxt)
: v(_v), nxt(_nxt) {}
... | ### Prompt
Generate a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 99;
struct edge {
int to, nxt;
} e[maxn << 1];
int head[maxn], tot;
void add(int x, int y) {
e[++tot].to = y;
e[tot].nxt = head[x];
head[x] = tot;
}
int f[maxn][30], b[maxn], n, siz, top, stac[maxn], g[maxn], fa[maxn];
long long ans = 0;
vecto... | ### Prompt
Please create a solution in Cpp to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const long long N = 3e5 + 5;
long long n;
vector<long long> G[N];
long long dp[N][20][2];
long long lev[N];
void dfs(long long u, long long p) {
vector<long long> ver;
for (long long i = 0; i < G[u].size(); ++i) {
long long v = G[u][i];
if (v == p) continue;
... | ### Prompt
Generate a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long inf = 1e9 + 7;
const long long max_n = 3e5 + 3;
long long n;
vector<long long> gr[max_n];
vector<long long> dp[max_n];
long long mx_k[max_n];
void dfs(long long v, long long pr) {
... | ### Prompt
Please provide a cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
int n;
const int MaxN = 3e5;
vector<int> adj[MaxN];
vector<int> prec[MaxN];
void dfs(const int u, const int p) {
prec[u] = {1, n};
adj[u].erase(remove(adj[u].begin(), adj[u].end(), p), adj[u].end());
for (auto &&v : adj[u]) {
dfs(v, u);
prec[u][0] = max(prec[u... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, bzmax = 20;
int n;
int h[N];
int f[N][bzmax];
int buf[N], par[N];
vector<pair<int, int> > upd[N];
vector<int> g[N];
long long res;
int dres;
inline void init() {
scanf("%d", &n);
for (int i = 1, u, v; i < n; i++) {
scanf("%d%d", &u, &v);
g... | ### Prompt
Your task is to create a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
vector<int> g[300005];
int dp[300005][25], dp2[300005][25];
int D[300005];
int n;
void dfs(int u, int p) {
D[u] = 1;
for (int i = 0; i < (int)(g[u]).size(); i++) {
int v = g[u][i];
if (v != p) {
dfs(v, u);
D[u] = max(D[u], D[v] + 1);
for (int d... | ### Prompt
Please create a solution in CPP to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
long long ans = 0;
int n, f1[300010], dp[300010][20], fa[300010], buf[300010], dep[300010];
vector<int> v[300010];
vector<pair<int, int> > upd[300010];
void dfs(int np, int fath) {
f1[np] = 1;
fa[np] = fath;
for (int &x : v[np]) {
if (x == fath) continue;
dfs(... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
const int C = 600001, D = 70;
vector<vector<int> > tr(C);
int ij[C], a, b, n, dp[C], ih[C], bst[C], maxson[C];
long long wynn = 0;
int pom[C];
void Sebasort(int tab[], int l, int r) {
int lim = 2, limi = l + 1, limj = l + 2, j = limi, i = l, k = l;
while (lim / 2 <= r -... | ### Prompt
In Cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> v[300007];
int lvl[300007];
int cnt[300007];
int f[300007][21];
int dp[300007][21];
long long ans = 0;
void dfs(int vertex, int prv) {
int sz = v[vertex].size();
lvl[vertex] = 1;
for (int i = 0; i < sz; ++i) {
int h = v[vertex][i];
if (h == ... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
long long ans;
int ne[N << 1], fi[N], zz[N << 1], tot, x, y, dp[N][25], T, n, Max[N],
dp2[N][25];
void jb(int x, int y) {
ne[++tot] = fi[x];
fi[x] = tot;
zz[tot] = y;
}
int cmp(int x, int y) { return dp[x][T] > dp[y][T]; }
void dfs(int x, int... | ### Prompt
Your task is to create a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
const int sz = 3e5 + 10, x = 5;
vector<int> sv[sz];
int n, k = 1, dp[sz], dp2[sz][x + 1], dp3[sz][x + 1];
long long an = 0;
int dfs(int v, int pr) {
int re = 0;
dp[v] = 1;
vector<int> sp;
for (int a = 0; a < sv[v].size(); a++) {
int ne = sv[v][a];
if (ne != ... | ### Prompt
Please provide a cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
inline bool C(int a, int b) { return a > b; }
inline void inc(int& a, const int& b) {
if (a < b) a = b;
}
struct d {
d* n;
int t;
d(d* a, int b) : n(a), t(b) {}
} * G[300005];
void I(int f, int t) { G[f] = new d(G[f], t); }
int A[300005], M[300005][20], F[300005][20], V[300005], n;
long... | ### Prompt
Construct a cpp code solution to the problem outlined:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of i... |
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f, oo = inf;
inline long long read() {
register long long x = 0, f = 1;
register char c = getchar();
for (; !isdigit(c); c = getchar())
if (c == '-') f = -1;
for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return ... | ### Prompt
In cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
long long ans;
const int C = 20;
vector<int> adj[300010];
int dp[C + 2][300010];
int mx[C + 2][300010];
int h[300010];
int n;
void dfs(int root, int dad = -1) {
h[root] = 1;
for (auto x : adj[root])
if (x != dad) {
dfs(x, root);
h[root] = max(h[root], h[... | ### Prompt
Create a solution in Cpp for the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ... |
#include <bits/stdc++.h>
using namespace std;
vector<int> g[300005];
int dp[300005][25];
int D[300005];
int n;
void dfs1(int u, int p) {
D[u] = 1;
for (int i = 0; i < (int)(g[u]).size(); i++) {
int v = g[u][i];
if (v != p) {
dfs1(v, u);
D[u] = max(D[u], D[v] + 1);
}
}
for (int d = 2; d <... | ### Prompt
Please create a solution in CPP to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long inf_ll = 1e18;
const int N = 3e5 + 5;
long long d[N], pt[N], ans, sum;
vector<long long> adj[N], dp[N], s;
multiset<long long> f[N];
void dfs1(int i, int p) {
pt[i] = d[i] = adj[i].size() - (i != 0);
dp[i].resize(d[i]);
for (auto& ... | ### Prompt
In Cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int const nmax = 300000;
vector<int> g[1 + nmax];
vector<pair<int, int>> dp[1 + nmax];
int n;
int extract(int node, int l) {
int pos = 0;
for (int jump = 16; 0 < jump; jump /= 2)
if (pos + jump < dp[node].size() && dp[node][pos + jump].second <... | ### Prompt
Construct a Cpp code solution to the problem outlined:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of i... |
#include <bits/stdc++.h>
using namespace std;
bool home = 1;
signed realMain();
signed main() {
home = 0;
if (home) {
freopen("input", "r", stdin);
} else {
ios::sync_with_stdio(0);
cin.tie(0);
}
realMain();
}
const long long N = (long long)3e5 + 7;
const long long M = 20;
long long n, dep[N], ret... | ### Prompt
Please provide a CPP coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n;
vector<int> adj[N];
int h[N];
int f[N][20], g[N][20];
void dfs(int u, int p) {
for (int v : adj[u])
if (v != p) {
dfs(v, u);
h[u] = max(h[u], h[v] + 1);
}
f[u][1] = g[u][1] = n;
for (int i = 2; i <= 19; ++i) {
int ... | ### Prompt
Please provide a cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkmax(T &a, T b) {
return (a < b) ? a = b, 1 : 0;
}
template <typename T>
bool chkmin(T &a, T b) {
return (a > b) ? a = b, 1 : 0;
}
inline int read() {
int x = 0, fh = 1;
char ch = getchar();
for (; !isdigit(ch); ch = getchar())
if ... | ### Prompt
Construct a CPP code solution to the problem outlined:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of i... |
#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int fa[N], f[N], g[N], sn[N], id[N], n, u, v;
vector<int> s[N], e[N], tmp[N];
long long ans, sum;
int clk;
int dfs(int u) {
int mx = 1;
id[++clk] = u;
for (auto v : e[u])
if (v != fa[u]) {
fa[v] = u;
mx = max(mx, dfs(v) + 1);
... | ### Prompt
Please create a solution in Cpp to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int NMAX = 3e5 + 10, LGMAX = 19;
class Reader {
public:
Reader() : m_pos(kBufferSize - 1), m_buffer(new char[kBufferSize]) { next(); }
Reader& operator>>(int& value) {
value = 0;
while (current() < '0' || current() > '9') next();
while (current() >= '... | ### Prompt
Please provide a cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
const int mod = 1e9 + 7;
long long ans;
int n, d[N], son[N], size[N];
vector<int> G[N], dp[N];
bool cmp(int i, int j) { return d[i] > d[j]; }
void dfs(int x, int f) {
size[x] = 1;
if (find(G[x].begin(), G[x].end(), f) != G[x].end())
G[x].erase... | ### Prompt
In cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
int m, k;
long long ans;
const int N = 330000;
int h[N], dp[N], mdp[N], cnt[N], par[N], chd[N];
vector<int> adj[N];
vector<int> V;
void dfs(int u, int p) {
V.push_back(u);
par[u] = p;
for (int v : adj[u]) {
if (v == p) continue;
dfs(v, u);
}
chd[u] = adj[u... | ### Prompt
Please formulate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n;
vector<int> adj[N];
vector<int> karr[N];
multiset<int> ss[N];
vector<int> *val[N];
long long *sum[N];
long long ans = 0;
void dfs(int v, int p) {
int c = adj[v].size() - (p > 0);
for (auto it : adj[v]) {
if (it == p) continue;
dfs(i... | ### Prompt
Please provide a CPP coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
inline bool C(int a, int b) { return a > b; }
inline void inc(int &a, const int &b) {
if (a < b) a = b;
}
struct d {
d *n;
int t;
d(d *a, int b) : n(a), t(b) {}
} * G[300005];
void I(int f, int t) { G[f] = new d(G[f], t); }
int A[300005], M[300005][20], F[300005][20], V[300005], n;
long... | ### Prompt
Generate a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
struct EDGE {
int to, next;
} e[600010];
int n, head[300010], top;
void add(int u, int v) {
e[top].to = v;
e[top].next = head[u];
head[u] = top++;
}
int dp[300010][20], fa[300010], dep[300010];
void dfs(int x) {
dp[x][1] = n;
int i, j;
for (i = head[x]; ~i; i ... | ### Prompt
In cpp, your task is to solve the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its... |
#include <bits/stdc++.h>
using namespace std;
int mx[300005], f[300005][20], n;
long long ans;
vector<int> e[300005];
int main() {
scanf("%d", &n);
for (int i = 1, u, v; i < n; i++)
scanf("%d%d", &u, &v), e[u].emplace_back(v), e[v].emplace_back(u);
function<void(int, int)> dfs1 = [&](int u, int fa) {
mx[u... | ### Prompt
Generate a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 300000;
const int MAXDEP = 18;
int n;
vector<int> adj[MAXN];
int par[MAXN];
int top[MAXN], ntop;
void dfsinit(int at) {
top[ntop++] = at;
for (int i = (0); i < (((int)(adj[at... | ### Prompt
Please provide a cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
struct edge {
int to, next;
} e[300005 * 2];
int head[300005], tot;
void add(int x, int y) {
e[++tot] = (edge){y, head[x]};
head[x] = tot;
}
int mx[300005], dp[300005][25];
long long ans;
int n, q[300005];
bool cmp(int x, int y) { return x > y; }
void dfs(int x, int f... | ### Prompt
Generate a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
std::vector<int> node[524288];
int mk[524288][32], km[524288][32];
int dpmk[524288][32], dpkm[524288][32];
long long int ans;
int a[524288];
int na;
int n, k, m;
void dfs(int p, int u) {
0;
for (auto v : node[u])
if (v != p) dfs(u, v);
for (int i = 1; i < k; ++i) {
na = 0;
for... | ### Prompt
Create a solution in Cpp for the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ... |
#include <bits/stdc++.h>
using namespace std;
vector<int> g[300010], ch[300010];
vector<pair<int, int> > up[300010];
int p[300010], dp[300010][30], val[300010];
int u, v, n;
long long ans;
void dfs1(int u, int f) {
p[u] = f;
for (int v : g[u])
if (v != f) {
ch[u].emplace_back(v);
dfs1(v, u);
}
}... | ### Prompt
Please provide a CPP coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
const int lg = 21;
int n, dp[maxn][lg], k, mx[maxn], d[lg];
long long s;
vector<int> ad[maxn];
void fix(int u, int p) {
for (auto it = ad[u].begin(); it != ad[u].end(); ++it) {
if (*it == p) {
ad[u].erase(it);
break;
}
}
f... | ### Prompt
Generate a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 100;
vector<int> G[N];
int dp[N][75];
int sub[N][75];
int dp1[N], dp2[N];
int M, n;
long long ans;
void dfs(int u, int p) {
int child = 0;
vector<int> ch_sz;
for (int v : G[u]) {
if (v == p) continue;
dfs(v, u);
for (int k = 1; k < M; k... | ### Prompt
Create a solution in Cpp for the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ... |
#include <bits/stdc++.h>
using namespace std;
inline char gc() {
static char buf[100000], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
? EOF
: *p1++;
}
inline int read() {
int x = 0;
char ch = getchar();
bool positive = 1;
for ... | ### Prompt
Please create a solution in CPP to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
struct edge {
int to, next;
} e[300005 * 2];
int head[300005], tot;
void add(int x, int y) {
e[++tot] = (edge){y, head[x]};
head[x] = tot;
}
int mx[300005], dp[300005][25], pp[300005][25];
long long ans;
int n, q[300005];
bool cmp(int x, int y) { return x > y; }
void ... | ### Prompt
Construct a Cpp code solution to the problem outlined:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of i... |
#include <bits/stdc++.h>
using namespace std;
const int NMAX = 3e5 + 10, LGMAX = 19;
int N;
int MaxLevel;
int level[NMAX], height[NMAX];
int64_t answer;
int DP[LGMAX][NMAX], father[NMAX], val[NMAX];
vector<int> T[NMAX], LevelNodes[NMAX];
vector<pair<int, int>> GoUp[NMAX];
int DFS(int node, int from, int level) {
fath... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n, cnt, last[N], dep[N], f[N][25], a[N], dp[N], fa[N];
struct edge {
int to, next;
} e[N * 2];
vector<pair<int, int> > vec[N];
long long ans;
void addedge(int u, int v) {
e[++cnt].to = v;
e[cnt].next = last[u];
last[u] = cnt;
e[++cnt].to ... | ### Prompt
Develop a solution in Cpp to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 5, MAXLOG = 20;
template <typename _T>
void read(_T &x) {
x = 0;
char s = getchar();
int f = 1;
while (s > '9' || s < '0') {
if (s == '-') f = -1;
s = getchar();
}
while (s >= '0' && s <= '9') {
x = (x << 3) + (x << 1) + (s - '... | ### Prompt
Your task is to create a Cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
bool mini(T1 &a, T2 b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <typename T1, typename T2>
bool maxi(T1 &a, T2 b) {
if (a < b) {
a = b;
return true;
}
return false;
}
const int N = 3e5 + 5;... | ### Prompt
Generate a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 300005, S = 550;
int n;
vector<int> g[N];
int p0[N];
int e[N];
vector<int> v0;
void dfs0(int x, int p) {
e[x] = 1;
p0[x] = p;
for (int i = 0; i < g[x].size(); ++i) {
... | ### Prompt
Develop a solution in Cpp to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int logn = 20;
int n;
vector<vector<int> > g, dp;
vector<int> f, maxd;
void dfs1(int v = 0, int p = -1) {
for (int i : g[v]) {
if (i != p) {
dfs1(i, v);
maxd[v] = max(maxd[i] + 1, maxd[v]);
}
}
dp[v][0] = n;
for (int l = 1; l < logn; l++) {... | ### Prompt
Please provide a cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
int N;
vector<int> graph[300005];
int mdp[20][300005];
int cdp[20][300005];
int mdep[300005];
int fre[300005];
long long ans = 0;
void dfs(int n) {
cdp[1][n] = N;
mdep[n] = 1;
for (int e : graph[n]) {
graph[e].erase(find(graph[e].begin(), graph[e].end(), n));
... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
template <typename T>
bool cmax(T &a, T b) {
return (a < b) ? a = b, 1 : 0;
}
template <typename T>
bool cmin(T &a, T b) {
return (a > b) ? a = b, 1 : 0;
}
template <typename T>
T read() {
T ans = 0, f = 1;
char ch = getchar();
while (!isdig... | ### Prompt
Construct a cpp code solution to the problem outlined:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of i... |
#include <bits/stdc++.h>
namespace fuck {
const int N = 300100, M = 20;
int begin[N], next[N * 2], to[N * 2];
int n, e, lim;
void add(int x, int y, bool k = 1) {
to[++e] = y;
next[e] = begin[x];
begin[x] = e;
if (k) add(y, x, 0);
}
void initialize() {
scanf("%d", &n);
for (int i = 1, u, v; i < n; i++) scanf... | ### Prompt
Your task is to create a cpp solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at lea... |
#include <bits/stdc++.h>
using namespace std;
const int LG = 20;
int dp[300100][LG], mx[300100][LG], dep[300100];
int buf[300100];
vector<int> g[300100];
long long solve(int u, int p) {
long long ans = 0;
dep[u] = 1;
for (auto v : g[u]) {
if (v == p) continue;
ans += solve(v, u);
dep[u] = max(dep[u], ... | ### Prompt
Generate a CPP solution to the following problem:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its ch... |
#include <bits/stdc++.h>
using namespace std;
template <class _Tp>
_Tp gcd(_Tp a, _Tp b) {
return (b == 0) ? (a) : (gcd(b, a % b));
}
const long long Inf = 1000000000000000000ll;
const int inf = 1000000000;
char buf[1 << 25], *p1 = buf, *p2 = buf;
inline int getc() {
return p1 == p2 && (p2 = (p1 = buf) + fread(buf,... | ### Prompt
Develop a solution in Cpp to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
long long read() {
char ch = getchar();
long long x = 0;
int op = 1;
for (; !isdigit(ch); ch = getchar())
if (ch == '-') op = -1;
for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
return x * op;
}
int n, cnt, head[300005], mx[300005], d... | ### Prompt
Develop a solution in CPP to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k o... |
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n;
vector<int> adj[N];
vector<int> karr[N];
multiset<int> ss[N];
vector<int> *val[N];
long long sum[N];
long long ans = 0;
void dfs(int v, int p) {
int c = adj[v].size() - (p > 0);
for (auto it : adj[v]) {
if (it == p) continue;
dfs(it... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at... |
#include <bits/stdc++.h>
using namespace std;
using cat = long long;
unsigned long long MOD = 998244353LL;
unsigned long long pw(unsigned long long a, unsigned long long e) {
if (e <= 0) return 1;
unsigned long long x = pw(a, e / 2);
x = (x * x) % MOD;
if (e % 2 != 0) x = (x * a) % MOD;
return x;
}
unsigned l... | ### Prompt
In CPP, your task is to solve the following problem:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one of the... |
#include <bits/stdc++.h>
using namespace std;
struct node {
int t, next;
} a[200010];
long long DCXISSOHANDSOME[20][262144], inv[262145], fac[100010], ifac[100010],
f[100010], g[100010], ans;
int head[100010], size[100010], h[100010], n, k, tt, tot;
inline int rd() {
int x = 0;
char ch = getchar();
for (; c... | ### Prompt
Develop a solution in cpp to the problem described below:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one o... |
#include <bits/stdc++.h>
template <typename _Tp>
void read(_Tp &x) {
char ch(getchar());
bool f(false);
while (!isdigit(ch)) f |= ch == 45, ch = getchar();
x = ch & 15, ch = getchar();
while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
if (f) x = -x;
}
template <typename _Tp, typename... Args>
void... | ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exa... |
#include <bits/stdc++.h>
using std::lower_bound;
using std::max;
using std::min;
using std::random_shuffle;
using std::reverse;
using std::sort;
using std::swap;
using std::unique;
using std::upper_bound;
using std::vector;
void open(const char *s) {}
int rd() {
int s = 0, c, b = 0;
while (((c = getchar()) < '0' ||... | ### Prompt
Develop a solution in CPP to the problem described below:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one o... |
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int Mod = 998244353, rt = 3;
void add(int &x, int y) {
x += y;
if (x >= Mod) x -= Mod;
}
int Pow(int x, int e) {
int ret = 1;
while (e) {
if (e & 1) ret = 1ll * ret * x % Mod;
x = 1ll * x * x % Mod;
e >>= 1;
}
return ret... | ### Prompt
Please formulate a cpp solution to the following problem:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one o... |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
mt19937 rnd(239);
const long double pi = acos(-1.0);
const long long INF = 1e18 + 239;
const int BIG = 1e9 + 239;
const int M = 2e5 + 239;
const int T = (1 << 19);
const long long MOD = 998244353;
const long long root = 3;
const long long sub = 15... | ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one of t... |
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
char c, sg = 0;
while (c = getchar(), (c > '9' || c ... | ### Prompt
Construct a CPP code solution to the problem outlined:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one of t... |
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int MOD = 998244353;
const int g = 3;
inline int Mul(int a, int b) {
unsigned long long x = (long long)a * b;
unsigned xh = (unsigned)(x >> 32), xl = (unsigned)x, d, m;
asm("divl %4;\n\t" : "=a"(d), "=d"(m) : "d"(xh), "a"(xl), "r"(MOD));
... | ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly... |
#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(228);
const int M = 998244353;
const int P = 15311432;
const int B = (1 << 23);
const int N = 1e6 + 7;
const int K = 18;
int w[K][N];
int fact[N];
int rev_fact[N];
int sz[N];
int dp[N];
vector<int> g[N];
vector<int> ans[N];
vector<int> res[N];
inline int add(int... | ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one of t... |
#include <bits/stdc++.h>
using namespace std;
namespace Base {
const int inf = 0x3f3f3f3f, INF = 0x7fffffff;
const long long infll = 0x3f3f3f3f3f3f3f3fll, INFll = 0x7fffffffffffffffll;
template <typename T>
void read(T &x) {
x = 0;
int fh = 1;
double num = 1.0;
char ch = getchar();
while (!isdigit(ch)) {
... | ### Prompt
In CPP, your task is to solve the following problem:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one of the... |
#include <bits/stdc++.h>
using namespace std;
namespace gbd_ns {
template <typename C>
struct is_iterable {
template <class T>
static long check(...);
template <class T>
static char check(int, typename T::const_iterator = C().end());
enum {
value = sizeof(check<C>(0)) == sizeof(char),
neg_value = size... | ### Prompt
Construct a CPP code solution to the problem outlined:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one of t... |
#include <bits/stdc++.h>
using namespace std;
int get() {
char ch;
while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
;
if (ch == '-') {
int s = 0;
while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
return -s;
}
int s = ch - '0';
while (ch = getchar(), ch >= '0' &... | ### Prompt
Generate a cpp solution to the following problem:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one of these ... |
#include <bits/stdc++.h>
using namespace std;
namespace NTT {
int mbase, base, root;
int w[1 << 19];
int rev[1 << 19];
int modPow(int b, int e) {
int r = 1;
while (e > 0) {
if (e & 1) r = ((long long int)r * b) % 998244353;
e >>= 1;
b = ((long long int)b * b) % 998244353;
}
return r;
}
int setBase(i... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exa... |
#include <bits/stdc++.h>
using namespace std;
const int N = 262333, mod = 998244353, G = 3, inv2 = (mod + 1) >> 1;
int read() {
int ret = 0;
char c = getchar();
while (!isdigit(c)) c = getchar();
while (isdigit(c)) ret = ret * 10 + (c ^ 48), c = getchar();
return ret;
}
namespace Math {
int fac[N], ifac[N], i... | ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exa... |
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int MOD = 998244353;
const int g = 3;
int qpow(int x, int y = MOD - 2) {
int res = 1;
for (; y; y >>= 1, x = (long long)x * x % MOD)
if (y & 1) res = (long long)res * x % MOD;
return res;
}
int n, K, jc[N], njc[N];
void Prework() {
... | ### Prompt
Develop a solution in cpp to the problem described below:
You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one o... |
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