示例#1
0
文件: main.cpp 项目: lcavalcante/AA
int main(){
    int n, m;
    while(true){
        cin >> n >> m;
        if(n == 0 && m ==0) break;
        adj.assign(n,vi());
        visitados.assign(n,0);
        s.clear();
        for(int k = 0; k < m; k++){
            int principal, dependente;
            cin >> principal >> dependente;
            (adj.at(principal-1)).push_back(dependente-1);
        }
        for(int l = 0; l < n; l++){
            if(visitados[l] == 0){
                dfs(l);
            }
        }
        string result;
        for(int d = s.size()- 1 ; d > 0;d--){
            stringstream ss;
            ss << s.at(d);
            result += ss.str() + " ";
        }
        stringstream ss;
        ss << s.at(0);
        result += ss.str();
        cout << result << endl;
    }
}
示例#2
0
int main(int argc, char** argv)
{
  cin.sync_with_stdio(false);
  sieve(1000000);
  //sieve(10000000);
  int L, U;
  while (cin >> L >> U)
  {
    //cout << L << " " << U << endl;
    int min_dist = 1000000;
    int max_dist = -1;
    pair<int, int> closest;
    pair<int, int> furthest;

    for (int i = L; i <= U; ++i) // find all primes within range
    {
      if (is_prime(i))
      {
        _primes.push_back(i);
      }
    }
    if (_primes.size() < 2)
    {
      cout << "There are no adjacent primes.\n";
    }
    else
    {
      for (int i = 0; i < _primes.size()-1; ++i) // iterate through all primes
      {
        int dist = _primes.at(i+1) - _primes.at(i);
        //cout << "dist " << _primes[i] << " to " << _primes[i+1] << " = " << dist << endl;
        if (dist < min_dist)
        {
          closest.first = _primes[i];
          closest.second = _primes[i+1];
          min_dist = dist;
        }
        if (dist > max_dist)
        {
          furthest.first = _primes[i];
          furthest.second = _primes[i+1];
          max_dist = dist;
        }
        //cout << _primes[i] << " ";
      }
      cout << closest.first << "," << closest.second << " are closest, "
           << furthest.first << "," << furthest.second << " are most distant.\n";
    }
    _primes.clear();
  }
  return 0;
} 
示例#3
0
文件: main.cpp 项目: lcavalcante/AA
void dfs(int u){
    if(visitados.at(u) == 0){

        //printf("dfs em %d\n", u);
        visitados[u] = 1;

        for(int i = 0; i < adj.at(u).size(); i++){
            int vizinho = adj.at(u)[i];
            if(!visitados[vizinho]){
                dfs(vizinho);

            }
        }
        //printf("adicionando %d ao stack\n", u+1);
        s.push_back(u+1);

   }
}