Search is not available for this dataset
name stringlengths 2 112 | description stringlengths 29 13k | source int64 1 7 | difficulty int64 0 25 | solution stringlengths 7 983k | language stringclasses 4
values |
|---|---|---|---|---|---|
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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;
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | 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... | JAVA |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 < ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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;
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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, ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 =... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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) {
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 |
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... | JAVA |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | 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... | PYTHON |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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][... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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;... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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] == ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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++;
}
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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++;
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | 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)... | JAVA |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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]... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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;... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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++) {
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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);
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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;
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 == ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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++;
}
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | 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... | JAVA |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | 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[]... | JAVA |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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;
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 =... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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++;
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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++) {
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | 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... | JAVA |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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);
... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | 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... | JAVA |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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",... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | 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... | JAVA |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 = ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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.... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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 == ... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
638_C. Road Improvement | 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 work simulta... | 2 | 9 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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)) ||
... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | 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 ... | JAVA |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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]) {
... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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 ... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | /**
* 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... | JAVA |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | 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... | JAVA |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 |
/*
* 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.... | JAVA |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #!/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:
... | PYTHON |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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];
... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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 != ... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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(... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | 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(... | PYTHON3 |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | 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 ... | JAVA |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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)... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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]... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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 ... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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;
... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | 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... | JAVA |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 |
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(... | JAVA |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | 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])
... | PYTHON3 |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | 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]... | PYTHON |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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... | CPP |
666_A. Reberland Linguistics | 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 should know l... | 2 | 7 | #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';
... | CPP |
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