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// Noic - Ideia 4 | |
// Exercício 1 | |
// Complexidade: O(N^2) | |
// Por Leonardo Paes | |
#include <bits/stdc++.h> | |
using namespace std; | |
int v[1000100]; |
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// Noic - Ideia 4 | |
// Exercício 1 | |
// Complexidade: O(N) | |
// Por Leonardo Paes | |
#include <bits/stdc++.h> | |
using namespace std; | |
int v[1000100]; |
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// Noic - Ideia 4 | |
// Exercício 2 | |
// Complexidade: O(N) | |
// Por Leonardo Paes | |
#include <bits/stdc++.h> | |
using namespace std; | |
int v[3000000]; |
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// Noic - Iniciante - Semana 55 - Problema 2 | |
// O(N) | |
#include <bits/stdc++.h> | |
using namespace std; | |
int main(){ | |
//um representa a quantidade de 1's, e dois a quantidade de 2's |
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#include <bits/stdc++.h> | |
using namespace std; | |
const int MAXN = 1e5 + 100; | |
const long long mod = 1e9+7; | |
vector<int> grafo[MAXN]; |
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#include <bits/stdc++.h> | |
using namespace std; | |
const int maxn = 2e5+100; | |
int vet[maxn], resp[maxn], ind[maxn], pre[maxn], nex[maxn]; | |
struct nod{ | |
int v, d; |
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#include <bits/stdc++.h> | |
using namespace std; | |
int main(){ | |
int n, m; | |
cin >> n >> m; | |
vector<int> a(n), b(m); |
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#include <bits/stdc++.h> | |
using namespace std; | |
const int maxn = 5e2 + 100; // Número máximo de caracteres da string S. | |
int n, dp[maxn][maxn]; // Declaro N e a DP. | |
string s; // Declaro a string S. |
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// Os parametros da função build são: | |
// O índice do nó atual, o l e o r do intervalo representado pelo nó. | |
void build(int node, int l, int r){ | |
// Se o nó atual é uma folha: | |
// O valor desse nó é simplesmente vet[l], ou vet[r], e então retorno a função. | |
if(l==r){ | |
tree[node]=vet[l]; | |
return; | |
} | |
// Caso contrário, declararemos mid como o índice que divide o nó atual em duas metades. |
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// Os parametros da função Update são: | |
// O nó atual, tl e tr, o intervalo que o nó atual abrange, idx e x, o índice que queremos atualizar e o seu valor. | |
void update(int node, int tl, int tr, int idx, int x){ | |
// Se o nó atual é uma folha, então ele representa o intervalo de idx até idx. | |
// Então, atualizaremos o seu valor e retornaremos a função. | |
if(tl==tr){ | |
tree[node]=x; | |
vet[l]=x; | |
return; | |
} |
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