prompt string | response string |
|---|---|
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 23;
const int MOD = 1e9 + 7;
const int SZ = 2e5 + 100;
vector<pair<int, int> > G[SZ];
vector<int> ans[SZ];
int nr[SZ];
void dfs(int v, int par) {
int p = 0;
for (auto& it : G[v]) {
int u = it.first, id = it.second;
if (u == par) continue;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.*;
import java.util.*;
/*
*/
public class C {
static FastReader sc=null;
static int days[];
public static void main(String[] args) {
sc=new FastReader();
PrintWriter out=new PrintWriter(System.out);
int n=sc.nextInt();
Node nodes[]=new Node[n];
for(int i=0;i<n;i++)nodes[i]=new Node... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<vector<pair<int, int> > > g(200100);
bool used[200100];
int wayChet[200100];
int wayTime[200100];
vector<vector<int> > res(200100);
void dfs(int j, int last) {
used[j] = 1;
int time = 1;
if (last == 1) time = 2;
if (last == 0) time = 1;
for (int i = 0; i < ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int n, k, col[maxn];
vector<pair<int, int> > g[maxn];
vector<int> d[maxn];
void dfs(int v, int p) {
int mx = 1;
for (int i = 0; i < g[v].size(); i++) {
if (g[v][i].first != p) {
if (mx == col[v]) mx++;
col[g[v][i].first] = mx;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
vector<pair<int, int> > adj[N];
vector<int> ans[N];
void dfs(int u, int p, int idx) {
int cnt = 0;
for (auto it : adj[u])
if (it.first != p) {
++cnt;
cnt += (cnt == idx);
ans[cnt].push_back(it.second);
dfs(it.first, u, ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
struct Valence {
int index;
int valence;
bool operator<(Valence const &b) const { return valence < b.valence; }
};
struct Edge {
int from;
int to;
};
struct EdgeId {
int edge;
int id;
bool operator<(EdgeId const &b) const { return edge < b.edge; }
bool ope... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<int> ans[200005];
vector<pair<int, int> > g[200005];
int n, m, k, i, j, a[200005], b[200005], res;
set<int> st;
void dfs(int v, int p, int c) {
int cur = 1;
for (int j = 0; j < g[v].size(); j++)
if (g[v][j].first != p) {
if (cur == c) cur++;
res =... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<vector<int> > g;
bool used[1234567];
map<pair<int, int>, int> mapka;
void dfs(int v, vector<vector<int> >& ans, int p = -1) {
used[v] = 1;
int cur = 0;
for (int i = 0; i < g[v].size(); ++i) {
int to = g[v][i];
if (!used[to]) {
if (cur > p) {
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... |
import java.io.*;
import java.math.BigInteger;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.StringTokenizer;
/**
* Created by Leonti on 2016-03-20.
*/
public class C {
public static void main(String[] args) {
InputReader inputReader = new InputReader(System.in);
Pr... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " = " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(na... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import os,sys
n = int(raw_input())
nbs = [[] for i in xrange(n+1)]
t=sys.stdin.read().split("\n")
for i in range(n-1):
l=t[i]
u,v=map(int,l.split())
nbs[u].append((v,i+1))
nbs[v].append((u,i+1))
left,right = 0,1
ans = [[] for i in xrange(n+1)]
total = 0
stateL = [[1,1,0] for i in xrange(1,n+1)]
while le... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> g[200007];
bool visited[200007];
int reservedDay[200007];
int res[200007];
vector<int> printRes[200007];
int maxDay = 0;
void visit(int u) {
visited[u] = true;
int avDay = 1;
for (int i = 0; i < g[u].size(); i++) {
if (!visited[g[u][i].secon... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
void bfs(vector<vector<uint64_t>>& v,
map<tuple<uint64_t, uint64_t>, uint64_t>& road_nr) {
uint64_t len = v.size();
vector<bool> visited(len, false);
visited[0] = true;
visited[1] = true;
deque<tuple<uint64_t, uint64_t>> p;
map<uint64_t, vector<uint64_t... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #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 = 2 * 200000;
const int MAXM = MAXN - 1;
int n;
int deg[MAXN];
int ghead[MAXN], gprv[2 * MAXM], gnxt[2 * MAXM], gto[2 * MAXM];
int ihead[MAXN], iprv[2 * MAXM], inxt[2 * MAXM];
int ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<pair<int, int>> g[n + 1];
for (int i = 1; i <= n - 1; ++i) {
int u, v;
cin >> u >> v;
g[u].emplace_back(v, i);
g[v].emplace_back(u, i);
}
int sum = 0;
vector<vect... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int mn = 1;
vector<vector<pair<int, int>>> G;
vector<vector<int>> day;
vector<int> used;
void DFS(int n, int cl) {
int cc = 1;
for (int i = 0; i < G[n].size(); i++) {
if (cl == cc) {
cc++;
}
if (!used[G[n][i].first]) {
day[cc - 1].push_back(G[n][... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.writ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long n;
vector<pair<long long, long long> > g[200005];
long long sz;
vector<long long> ans[200005];
void dfs(long long s, long long pr, long long tt) {
long long cur = 0;
for (auto i : g[s]) {
if (i.first == pr) continue;
cur++;
if (cur == tt) cur += 1;... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<long long> v[200001];
long long ans = 0;
vector<long long> m[200001];
map<pair<long long, long long>, long long> edges;
void dfs(long long x, long long p, long long w) {
long long d = 1;
for (long long i = 0; i < (long long)v[x].size(); i++) {
if (v[x][i] == ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MM = 2e5 + 127;
vector<pair<int, int> > edge[MM];
int degree[MM];
vector<int> ans[MM];
int maxT;
void dfs(int cur, int par, int when) {
int timer = 0;
for (auto u : edge[cur]) {
if (u.first != par) {
if (timer == when) {
timer++;
}
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 200100, mod = 1e9 + 7, maxa = 1e6 + 100, maxb = 23;
const long long inf = 2e18 + 13;
long long max(long long x, long long y) { return (x > y ? x : y); }
long long min(long long x, long long y) { return (x < y ? x : y); }
vector<pair<int, int> > e;
vector<in... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
const double pi = 3.14159265359;
const int MOD = 1000000007;
const int dr[4] = {-1, 1, 0, 0};
const int dc[4] = {0, 0, -1, 1};
double sinA(double dig) { return sin(dig * pi / 180); }
double cosA(double dig) { return cos(dig * pi / 180); }
double tanA(double dig) {
int we = dig;
if (we == di... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
struct Edge {
int to;
int index;
int color;
size_t inv_pos;
Edge(int to = -1, int index = 0, size_t inv_pos = 0)
: to(to), index(index), color(-1), inv_pos(inv_pos) {}
};
int n;
vector<vector<Edge>> g;
void bfs(int start) {
queue<pair<int, int>> q;
q.pus... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MN = 2e5 + 44;
vector<pair<int, int> > graph[MN];
vector<int> ans[MN];
void dfs(int x, int y = -1, int color = -1) {
int last = 0;
for (auto v : graph[x])
if (v.first != y) {
if (last == color) last++;
ans[last].push_back(v.second);
dfs(v... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<int> gr[200500], v[200500];
map<pair<int, int>, int> mp;
int c[200500];
void dfs(int v, int par, int last) {
int col = 1;
if (last == col) col++;
for (int x : gr[v]) {
if (x == par) continue;
c[mp[make_pair(v, x)]] = col;
dfs(x, v, col);
col++;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.*;
import java.util.*;
import java.math.*;
import java.math.BigInteger;
public final class C
{
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans=new StringBuilder();
static FastReader in=new FastReader();
static ArrayList<ArrayList<edge>> g;
static long mod=(long)... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<int> adj[200004], ans[200004];
int clr[200004], mx = 0;
map<int, map<int, int> > mp;
void bfs(int src) {
int u, v, p, q, r, sz;
queue<int> qu;
qu.push(src);
while (!qu.empty()) {
u = qu.front();
qu.pop();
sz = adj[u].size();
p = 0;
for (in... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int nmax = 300010;
int n;
vector<pair<int, int> > g[nmax];
int u[nmax];
int color[nmax], cnt[nmax];
vector<int> ans[nmax];
void dfs(int v, int p = -1, int come = 0) {
int ptr = 1;
for (int i = 0; i < g[v].size(); i++) {
int num = g[v][i].second, to = g[v][i].f... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MaxN = 2e5, MaxM = 4e5;
int n, all;
int pre[MaxM + 5], last[MaxN + 5], other[MaxM + 5];
bool vis[MaxN + 5];
int ans[MaxM + 5], num[MaxM + 5], used[MaxN + 5];
int seq[MaxM + 5];
vector<int> v[MaxN + 5];
void Build(int x, int y, int d) {
pre[++all] = last[x];
la... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int cunt[200000 + 342];
bool printed[200000 + 342];
vector<pair<int, int> > adj[200000 + 342];
int pussy[200000 + 342];
vector<int> bc[200000 + 342];
void solve(int cur, int par, int t) {
int now = 0;
for (int i = 0; i < adj[cur].size(); i++) {
int nxt = adj[cur][i]... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int read() {
int ans = 0, f = 1;
char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
ans = ans * 10 + c - '0';
c = getchar();
}
return ans * f;
}
const int N = 2e5 + 5;
int n, d... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
const int INF = 0x3f3f3f3f;
int n, du[MAXN];
struct Edge {
int to, next;
} G[MAXN << 1];
int sz, head[MAXN];
void add(int u, int v) {
G[sz] = {v, head[u]};
head[u] = sz++;
}
vector<int> ret[MAXN];
void dfs(int u, int fa, int last) {
int now... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = (int)2e5 + 5;
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const long long LLINF = (long long)1e18;
int n;
vector<pair<int, int> > g[N];
vector<int> roads[N];
void dfs(int v, int p = -1, int id = -1) {
int pos = -1;
for (auto t : g[v]) {
int... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int max_n = 2e5 + 100;
struct edge {
int to, num, color, rev, next;
} E[max_n << 1];
int head[max_n];
int cnt = 1;
void add(int from, int to, int num, int rev, int color = 0) {
E[cnt].to = to;
E[cnt].next = head[from];
E[cnt].num = num;
E[cnt].color = color;... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long fp(long long a, long long b) {
if (b == 0) return 1;
long long x = fp(a, b / 2);
x = (x * x) % mo... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<long long> vis(200010);
vector<long long> adj[200010];
vector<long long> res[200010];
map<pair<long long, long long>, long long> M;
long long n, groups = 0;
void dfs(long long node, long long col) {
vis[node] = 1;
for (int i = 0; i < adj[node].size(); i++) {
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
vector<pair<int, int> > g[MAXN];
int n;
bool read() {
if (scanf("%d", &n) < 1) {
return false;
}
for (int i = 0; i < (int)n; ++i) {
g[i].clear();
}
for (int i = 0; i < (int)n - 1; ++i) {
int a, b;
scanf("%d%d", &a, &b);
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long int node, edge, vis[200010], dist[200010], col[200010], in[200010],
out[200010], timer = 1;
vector<long long int> gh[200010], ind[200010], ans[200010];
void dfs(long long int current, long long int blocked) {
long long int cnt = 0;
vis[current] = 1;
for ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
struct Edge {
size_t to, id;
Edge(size_t t = 0, size_t i = 0) : to(t), id(i) {}
};
vector<Edge> edges[200005];
vector<size_t> ans[200005];
void dfs(size_t v, size_t p = -1, size_t day = -1) {
size_t currDay = 0;
if (currDay == day) ++currDay;
for (size_t i = 0; i ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
void fre() {
freopen("c://test//input.in", "r", stdin);
freopen("c://test//output.out", "w", stdout);
}
template <class T1, class T2>
inline void gmax(T1 &a, T2 b) {
if (b > a) a = b;
}
template <class T1, class T2>
inline void gmin(T1 &a, T2 b) {
if (b < a) a = b;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int M = 200005;
const int mod = 1e9 + 7;
map<int, int> id[M];
vector<int> adj[M], ans[M];
int n, mans;
void dfs(int node, int p, int l) {
mans = max(mans, (int)adj[node].size());
int turn = 1;
for (auto i : adj[node]) {
if (i == p) continue;
if (turn == ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > adjList[200010];
vector<int> day[200010];
void dfs(int u, int p, int sk) {
int currd = 0;
for (auto x : adjList[u]) {
int v = x.first;
if (v == p) continue;
if (currd == sk) currd += 1;
day[currd].push_back(x.second);
dfs(v, u... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int mn = 200010;
int n, k;
vector<int> ans[mn];
vector<pair<int, int> > e[mn];
int clr[mn];
void dfs(int v, int p, int cl) {
int q = 1;
for (int i = 0; i < e[v].size(); i++) {
if (e[v][i].first == p) {
continue;
}
if (q == cl) {
q++;
}
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.util.*;
import java.io.*;
public class TaskC {
private FastScanner in;
private PrintWriter out;
private class Edge {
int from;
int to;
int index;
public Edge(int from, int to, int index) {
this.from = from;
this.to = to;
this... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
public class C {
static ArrayList<Integer>[][] way;
static ArrayList<ArrayList<Integer>> answer;
static PrintWriter out;
public static void main(String[]... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 200000;
std::vector<int> ans[N];
std::vector<std::pair<int, int> > edge[N];
int maxi = 0;
void dfs(int u, int par, int day) {
int cnt = 0;
for (int i = 0; i < edge[u].size(); i++) {
std::pair<int, int> v = edge[u][i];
if (v.first == par) continue;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
using LL = long long;
using VI = vector<int>;
using VC = vector<char>;
using VS = vector<string>;
using VL = vector<long long>;
using VVI = vector<VI>;
using VVL = vector<VL>;
using MII = map<int, int>;
using MIVI = map<int, VI>;
using MSS = map<string, string>;
using MLL =... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = 200123;
vector<pair<int, int> > nbs[N];
int deg[N];
int act[N];
int taken[N];
int par[N];
int paredge[N];
int asparent[N];
int h[N];
int n, a, b;
void dfs(int v, int p = -1, int e = -1, int ch = 0) {
par[v] = p;
paredge[v] = e;
h[v] = ch;
for (pair<int... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int N = (int)2e5 + 1;
struct v {
int i, c;
};
int n, a, b;
map<int, map<int, int>> tr;
int que[N], l, r;
stack<int> st;
int bV[N] = {0};
v V[N] = {0};
bool d[N] = {0};
deque<deque<int>> de;
void rec(int i = V[0].i, int c = 1) {
for (map<int, int>::iterator it = tr... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<int> ans[200001];
vector<pair<int, int> > v[200001];
int vis[200001];
void dfs(int s, int r) {
int d = 0;
vis[s] = 1;
for (auto i : v[s]) {
if (vis[i.first] == 0) {
int a = i.first;
int b = i.second;
d++;
if (r == d) {
d++;
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
inline long long in() {
int32_t x;
scanf("%d", &x);
return x;
}
inline string getStr() {
char ch[1000000];
scanf("%s", ch);
return ch;
}
inline string getStr2() {
char ch[5];
scanf("%s", ch);
return ch;
};
template <class P, class Q>
inline P smin(P &a, Q ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int n, result = 0, lab[MAXN];
vector<pair<int, int> > adj_list[MAXN];
vector<vector<int> > ans;
void dfs(int u, int last, int skip) {
int curr;
if (skip != 1)
curr = 1;
else
curr = 2;
for (int i = 0; i < adj_list[u].size(); i++) {
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
ArrayList<Integer>[] g;
ArrayList<Integer>[] id;
ArrayList<Integer>[] at;
boolean[] was;
int[] par;
int[] parColor;
public void solve(int testNumber, FastScanner in, PrintWriter out) {
int n = i... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> e[200005];
vector<int> ans[200005];
int n, all;
void dfs(int fa, int u, int k) {
int day = 0;
for (auto i : e[u]) {
int v = i.first, id = i.second;
if (fa != v) {
if (++day == k) {
day++;
}
all = max(all, day);
... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import javafx.geometry.Pos;
import javafx.util.Pair;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.security.cert.PolicyNode;
import java.util.*;
public class Solution {
BufferedReader in;
PrintWriter out;
StringTokenizer st;
String nextToken() throws IOExc... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
long long gcd1(long long a, long long b) {
if (a == 0) return b;
return gcd1(b % a, a);
}
long long modx(long long base, long long ex) {
long long ans = 1LL, val = base;
while (ex > 0LL) {
if (ex & 1LL) ans = (ans * val) % 1000000009LL;
val = (val * val) % 1... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000 + 42;
struct {
int u, v;
int to(int from) const { return u + v - from; }
} edges[MAXN];
list<int> graph[MAXN];
list<int> roads[MAXN];
bool used[MAXN] = {};
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i < n; ++i) {
scanf("%d%d",... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
/**
* Created by roman on 21.03.2016.
*/
public class Task3 {
public BufferedReader bufferedReader;
public StringTokenizer stringTokenizer;
public PrintWriter print... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int n;
vector<pair<int, int> > g[200000];
int col[200000];
vector<int> plan[200000];
void dfs(int v, int p, int bad) {
int cur = 0;
for (pair<int, int> to : g[v])
if (to.first != p) {
if (cur == bad) ++cur;
col[to.second] = cur;
dfs(to.first, v, cu... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
typedef struct {
int sum, suf, pre, max;
} Node;
int toint(const string &s) {
stringstream ss;
ss << s;
int x;
ss >> x;
return x;
}
const int MAXN = 2e5 + 100;
const int UP = 31;
const long long int highest = 1e18;
const double pi = acos(-1);
const double Phi = ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
void bfs(vector<vector<uint64_t>>& g,
map<tuple<uint64_t, uint64_t>, uint64_t>& i2r) {
vector<bool> visit(g.size());
deque<tuple<uint64_t, uint64_t, uint64_t>> ev;
unordered_map<int64_t, vector<uint64_t>> roads_by_day;
visit[1] = true;
int day = 0;
for ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
vector<int> v[222222];
map<pair<int, int>, int> m;
bool vis[222222];
const long long INF = 1000000000;
vector<int> ans[222222];
int ans1 = 1;
void dfs(int k, int last) {
int num = 0;
for (int i = 0; i < v[k].size(); ++i)
if (!vis[v[k][i]]) {
if (num == last) n... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
int N, maxd;
vector<pair<int, int> > adj[200005];
vector<int> days[200005];
bool seen[200005];
void dostuff(int n, int d) {
seen[n] = true;
int i = 1;
for (auto a : adj[n]) {
if (!seen[a.first]) {
if (i == d) {
i++;
}
days[i].push_back(a.... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int mxn = (2e5) + 5;
vector<pair<int, int> > tr[mxn];
vector<int> ans[mxn];
int mx;
void dfs(int v, int p, int in) {
mx = max(mx, in + 1);
int c = -1;
for (int i = 0; i < tr[v].size(); ++i) {
int u = tr[v][i].first;
int x = tr[v][i].second;
if (u == ... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
const int md = 1e9 + 7;
const int INF = 1e9 + 7;
const double EPS = 1e-10;
int n;
const int MAXN = 200010;
map<pair<int, int>, int> edge;
vector<int> g[MAXN];
bool vis[MAXN];
int cant[MAXN];
vector<int> first_not_used(MAXN, 1);
vector<int> day[MAXN];
int ans;
void dfs(int v... |
Problem: In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to wor... | #include <bits/stdc++.h>
using namespace std;
class graphal {
public:
int n, mx = 0;
vector<pair<int, int> > *ed;
vector<int> *ans;
graphal(int n) {
this->n = n;
ed = new vector<pair<int, int> >[n];
ans = new vector<int>[n + 1]();
}
~graphal() {
delete[] ed;
delete[] ans;
}
void add... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
const int N = 10010;
long long t2[N], t3[N];
void solve() {
string s;
cin >> s;
int n = s.size();
s += "#####";
long long res = 0;
set<string> suffs;
t2[n - 2] = 1;
t3[n - 3] = 1;
for (int i = n - 1; i >= 5; --i) {
if (i + 3 < n) {
if (s.substr(i... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
template <class T, class U>
inline void Max(T &a, U b) {
if (a < b) a = b;
}
template <class T, class U>
inline void Min(T &a, U b) {
if (a > b) a = b;
}
inline void add(int &a, int b) {
a += b;
while (a >= 1000000007) a -= 1000000007;
}
int pow(int a, int b) {
in... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
long long int a[1000005][5] = {0};
map<string, long long int> mp, tot;
void fn(long long int l, long long int r);
string str;
int main() {
cin >> str;
fn(0, str.length() - 1);
cout << tot.size() << endl;
for (auto t : tot) cout << t.first << endl;
return 0;
}
void... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
char s[100111];
int n;
int F[100111][2];
int sz[2] = {2, 3};
int Can(int k, int tp) {
if (F[k][tp] != -1) return F[k][tp];
if (k + sz[tp] - 1 == n) return 1;
if (k + sz[tp] - 1 > n) return 0;
F[k][tp] = 0;
if (tp == 0) {
if (s[k] == s[k + 2] && s[k + 1] == s[k... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
int main() {
string s;
cin >> s;
int len = s.length();
bool dp[len + 2][2];
memset(dp, false, sizeof(dp));
dp[len][0] = dp[len][1] = true;
set<string> ans;
for (int i = len - 2; i > 4; --i) {
if ((dp[i + 2][0] && s.substr(i, 2) != s.substr(i + 2, 2)) ||
... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | import java.io.*;
import java.lang.reflect.Array;
import java.math.*;
import java.util.*;
public class icpc
{
static String s = "";
static HashSet<String> h = new HashSet<>();
static HashSet<Integer> h1 = new HashSet<>();
public static void main(String[] args) throws IOException
{
// Reader ... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
string s;
set<string> vals;
int N;
vector<bool> seen;
map<pair<pair<int, int>, int>, bool> rem;
bool cansplit(int current, int prevl, int length) {
pair<pair<int, int>, int> key = make_pair(make_pair(current, prevl), length);
map<pair<pair<int, int>, int>, bool>::iterat... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
bool valid[10000 + 100];
set<string> ans;
int main() {
string s;
getline(cin, s);
valid[s.length()] = 1;
for (int i = s.length() - 1; i > 4; --i) {
if (valid[i + 2]) {
string t = s.substr(i, 2);
if (s.find(t, i + 2) != i + 2 || valid[i + 5]) {
... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y) {
if (y == 0) return 1ll;
if (y % 2)
return (x % 1000000007 * power(x, y - 1) % 1000000007) % 1000000007;
else
return power((x * x) % 1000000007, y / 2) % 1000000007;
}
set<string> ans;
int main() {
ios_base::sync_wi... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100000;
char S[maxn];
set<string> ans;
bool f[maxn][3];
int main() {
scanf("%s", S);
int len = strlen(S);
f[len - 2][0] = true;
f[len - 3][1] = true;
for (int i = len - 2; i >= 5; --i) {
if (f[i + 2][0] && (S[i] != S[i + 2] || S[i + 1] != S[i ... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | /**
* Created by yume on 2016/5/25.
*/
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.TreeSet;
public class Main {
public static void main(final String[] args... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
const int MM = 1e4 + 5;
int n;
bool dp[MM][2];
string s;
int main() {
cin.tie(0)->sync_with_stdio(0);
cin >> s;
n = s.size();
if (n == 5) return cout << "0\n", 0;
set<string> ss;
dp[n - 2][0] = dp[n - 3][1] = 1;
for (int i = n - 4; i > 4; i--) {
string two... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
int n;
bool dp[10001][2];
string str;
set<string> s;
int main() {
cin >> str;
n = str.length();
for (register int i = n - 2; i >= 5; --i) {
if (i + 2 == n) dp[i][0] = 1;
if (i + 3 == n) dp[i][1] = 1;
if (dp[i + 3][0]) dp[i][1] = 1;
if (dp[i + 2][1]) dp... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | import java.util.Scanner;
import java.util.TreeSet;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
TreeSet<String> suffixes = new TreeSet<>();
String word = sc.nextLine();
sc.close();
String tempSubstring;
boolean... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... |
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #!/usr/bin/python2
# -*- coding: utf-8 -*-
import sys
import os
def build_must(s):
s = "".join(reversed(s))
r = [None, None, s[:2], s[:3]]
c3 = s[:3]
c2 = s[1:3]
for c in s[3:]:
c3 = c3[-2] + c3[-1] + c
c2 = c2[-1] + c
t2 = t3 = None
if r[-3] is not None:
... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
const int oo = (int)1e9;
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
set<string> seen[10009], ret;
string s;
int n;
void dfs(int ind, string next) {
if (!seen[ind].insert(next).second) return;
if (ind - 2 > 4) {
string x = "";
x += s[ind - 2];
... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
const double eps = 0.000000000000001;
string convertstring(long long n) {
stringstream ss;
ss << n;
return ss.str();
}
int len, dp[10005][4];
string s;
int solve(int ind, int last) {
if (ind == len) return dp[ind][last] = 1;
int &ans = dp[ind][last];
if (ans != ... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
string s;
set<string> tmp;
bool dp[10010][5];
void go(int i, int prv) {
if (dp[i][prv]) return;
dp[i][prv] = 1;
if (s.substr(i, 2) != s.substr(i - prv, prv) && i + 6 < s.size()) {
string t = s.substr(i, 2);
reverse(t.begin(), t.end());
tmp.insert(t);
g... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
string s;
unordered_set<string> suffixes;
bool checked[2][10001];
void solve(int end, const string& prev) {
bool& c = checked[prev.length() - 2][end];
if (c) return;
c = true;
for (int i = 2; i < 4; ++i) {
if (end - i < 5) continue;
string suffix = s.substr(... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | from sys import *
setrecursionlimit(200000)
d = {}
t = set()
s = input() + ' '
def gen(l, ll):
if (l, ll) in t: return
t.add((l, ll))
if l > 6:
d[s[l - 2 : l]] = 1
if s[l - 2 : l] != s[l : ll]: gen(l - 2, l)
if l > 7:
d[s[l - 3 : l]] = 1
if s[l - 3 : l] != s[l : ll]: gen(... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
public class Main {
private static StringTokenizer st;
private static BufferedReader br;
public static long MOD = 1000000007;
public ... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
x = 0;
char ch = getchar();
int fh = 1;
while (ch < '0' || ch > '9') {
if (ch == '-') fh = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
x *= fh;
}
template <typename... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #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;
}
int power(int a, int b, int m, int ans = 1) {
for (; b; b >>= 1, a = 1LL * a * a % m)
if (b & 1)... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
int dp[2][10005];
char s[10005];
int main() {
cin >> s;
int n = strlen(s);
dp[0][n - 2] = 1;
dp[1][n - 3] = 1;
for (int i = n - 4; i >= 0; i--) {
if (dp[1][i + 2] == 1 ||
(dp[0][i + 2] == 1 && (s[i] != s[i + 2] || s[i + 1] != s[i + 3])))
dp[0][i]... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
#pragma GCC optimize("-O3")
using namespace std;
using lli = long long int;
using llu = long long unsigned;
using pii = tuple<lli, lli>;
using piii = tuple<lli, lli, lli>;
using piiii = tuple<lli, lli, lli, lli>;
using vi = vector<lli>;
using vii = vector<pii>;
using viii = vector<piii>;
using ... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
int n;
string s;
set<string> a;
int dp[10010][2];
bool ok(int i, int x) {
if (i + x > n) return 0;
if (i + x == n) return 1;
if (i + x == n - 1) return 0;
if (dp[i][x - 2] != -1) return dp[i][x - 2];
int &ref = dp[i][x - 2];
if (ok(i + x, 5 - x))
ref = 1;
... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | import java.util.*;
import java.io.*;
public class a {
public static int[][] memo;
public static String s;
public static void main(String[] args){
Scanner br = new Scanner(System.in);
s = br.next();
memo = new int[4][s.length()];
for(int i = 0;i<4;i++){
Arrays.fill(memo[i], -1);
}
TreeSet<String> suff... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cer... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... |
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
public class MO {
public static void main(String[] args) throws NumberFormatException, IOException {
Scanner in = new Scanner(... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | import sys
sys.setrecursionlimit(10000)
s = input()
s = s[5:] + " "
res = set()
aux = set()
def getWords(x,y):
if (x,y) in aux:
return
aux.add((x,y))
if x > 1 and s[x:y] != s[x-2:x]:
res.add(s[x-2:x])
getWords(x-2,x)
if x > 2 and s[x:y] != s[x-3:x]:
res.add(s[x-3:x])
... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | s = raw_input().strip()
l = len(s)
a = [[] for i in xrange(l + 2)]
a[l-1] = a[l+1] = None
for i in xrange(l - 2, 4, -1):
t = s[i:i+2]
if a[i+2] is not None and a[i+2] != [t]:
a[i].append(t)
t = s[i:i+3]
if a[i+3] is not None and a[i+3] != [t]:
a[i].append(t)
if not a[i]:
a[i]... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 5;
int dp[N][4];
set<string> st[26];
set<string>::iterator it;
string s;
int n;
void solve(int idx, int last) {
if (idx <= 5) {
return;
}
if (last != 0 && dp[idx][last - idx] != 0) {
return;
}
string t, u;
t.push_back(s[idx]);
t.pus... |
Problem: First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you shou... | #include <bits/stdc++.h>
using namespace std;
template <typename Tp>
inline void read(Tp &x) {
static char c;
static bool neg;
x = 0, c = getchar(), neg = false;
for (; !isdigit(c); c = getchar()) {
if (c == '-') {
neg = true;
}
}
for (; isdigit(c); c = getchar()) {
x = x * 10 + c - '0';
... |
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