miguelAlessandro

//Bitset
#include <iostream>
using namespace std;

struct Bitset{
  long long m;
  Bitset(int x = 0){m = x;}
  const static long long empty_set = 0;
  bool Find(int in){return m&(1LL<<in);}
  void Erase(int in){ m &= ~in;}

miguelAlessandro

///Soluciones

1. hacer una función que bote -1 en caso que un entero sea negativo y 0 en otro caso.
rpta:

int neg(int x){return -(x < 0);}

2. intercambie dos enteros x e y, se entiende que ahora x sera y e y tendra el valor inicial de x. 
rpta:

miguelAlessandro

1. http://codeforces.com/contest/467/problem/B
2. http://codeforces.com/contest/535/problem/B
3. http://codeforces.com/problemset/problem/550/B

miguelAlessandro

• Segment Trees / let us code
• segment tree and lazy propagation / se7so
• generalizing segment trees
• Dynamic Segment Trees
• Modular Segment Tree with Lazy Propagation
• persistent segment tree / iq opengenus
• persistent segment trees explained with spoj problems / anudeep
• mit / persistent
• persistent segment tree / lyzqm
• [persistent segment tree / secmem](http://www.secm

miguelAlessandro

//fenwick_tree.cpp
//complexity:
//  update - O(log n)
//  query - O(log n)
//  sum_range - O(log n)
// binary_fenwick - O(log n)

const int maxN = 100002;
int ft[maxN];

miguelAlessandro

//sparse_table
//complexity:
//  min_range - O(1)
//  build - O(n * log n)

const int N = 100002;
const int LogN = 18;
int st[N][LogN];
int n;

miguelAlessandro

//segment tree
//complexity:
//  merge - O(1)
//  shift - O(1)
//  upd - O(1)
// build - O(n)
// update - O(log n)
//  query - O(log n)

const int maxN = 100002;

miguelAlessandro

//sqrt descomposition
//complexity:
//  build - O(n)
//  query - O(sqrt n)
//  update - O(1)

const int N = 100005;
int block[350], a[N];
int n, nb;

miguelAlessandro

const int N = 50002;
int len, n, q;
int vis[N], ans[N];
int a[N], ans_mo;

struct Query{int l, r, in;} Q[N];

bool cmp(const Query& p, const Query& q){
  return p.r/len < q.r/len or p.r/len == q.r/len and p.l > q.l;
}

miguelAlessandro

int times = 0;
int L[N], R[N], sz[N];
bool S1[N], S2[N];
int ans[N], p[N];

void dfs(int x, int p){
    L[x] = times++;
    for(int v : adj[x])
      if(v != p) dfs(v, x);
    R[x] = times++;