Eratosthens sieve

#include <iostream>
#include <set>

using namespace std;

#include "IO.hpp"

void SIEVE(set<int>& primes,int limit){

  int index;

  primes.erase(primes.begin(),primes.end());

  for(int i=2;i<limit;i++)
    primes.insert(i);

  for(int i=2;i*i<=limit;i++)
    if(primes.find(i) != primes.end()){

      index = 2*i;
      while(index <= limit){
        primes.erase(index);
        index += i;
      }
    }

  return;
}

int main(int argc,char *argv[])
{
  set<int> primeSet;

  set<int>::iterator iter;
  int primeLimit,count=0;

  cout<<"Enter upper limit for the set of primes: ";
  cin>>primeLimit;
  cout<<endl;

  SIEVE(primeSet,primeLimit);

  iter= primeSet.begin();

  writeContainer(primeSet.begin(),primeSet.end()," ",10);

  cout<<endl;

  return 0;
}

#ifndef _IO_HPP_
#define _IO_HPP_

#include <vector>
#include <set>
#include <list>

using namespace std;

//LINE=0 不需换行
//separator 表示每个元素间隔符号
//N为元素个数

//writeArray      (arr[],     N:,        separator,  LINE)
//writeContainer  (begin(),   end(),     separator,  LINE)
//writeVector     (vector<T>, separator, LINE)
//writeList       (list<T>,   separator, LINE)
//writeSet        (set<T>,    separator, LINE)



template <typename T>
void writeArray(const T arr[],                  \
                int N,                          \
                const char* separator,          \
                int LINE){

	for(int i=0;i<N;i++){

		cout<<arr[i]<<separator;

    if(LINE!=0)
      if(i%LINE==0)
        cout<<endl;
  }
	cout<<endl;

  return;
}

template <typename T>
void writeVector(const vector<T>& v,            \
                 const char* separator,         \
                 int LINE){

	int i, n = v.size();

	for(i=0;i<n;i++){

		cout<<v[i]<<separator;

    if(LINE!=0)
      if(i%LINE==0)
        cout<<endl;
  }
	cout<<endl;

  return;
}

template <typename T>
void writeList(const list<T>& alist,            \
               const char* separator,           \
               int LINE){

	typename list<T>::const_iterator iter;

  int count = 0;
  iter = alist.begin();

  while(iter != alist.end()){

    count++;
		cout<<*iter<<separator;

    if(LINE!=0)
      if(count%LINE==0)
        cout<<endl;

    iter++;
  }
  cout<<endl;

  return;
}

template <typename T>
void writeSet(const set<T>& alist,              \
              const char* separator,            \
              int LINE){

	typename set<T>::const_iterator iter;

  int count = 0;
  iter = alist.begin();

  while(iter != alist.end()){

    count++;
		cout<<*iter<<separator;

    if(LINE!=0)
      if(count%LINE==0)
        cout<<endl;

    iter++;
  }
  cout<<endl;

  return;
}

template <typename Iterator>
void writeContainer(Iterator begin,             \
                    Iterator end,               \
                    const char* separator,      \
                    int LINE){

	Iterator iter = begin;
  int count = 0;

	while (iter!=end){

    count++;
		cout<<*iter<<separator;

    if(LINE!=0)
      if(count%LINE==0)
        cout<<endl;

		iter++;
	}
  cout<<endl;

  return;
}

#endif

这个筛实现很简单, 比如上限是300 就把300以内所有小于sqrt(300)质数的倍数划掉就行, 我这里划了小于sqrt(300)所有数的倍数, 因为要划小于sqrt(300)的质数必须先生成, 用两此筛就行

比如筛10000以内  先生成100以内的质数 然后划掉100以内的质数就可以,不用划100个数字,会节约点时间



按照SPOJ上的提示,用筛法,二次筛 都无法通过SPOJ第二题, 都是超时.
或许我筛法写得性能不好 像set容器 性能可能确实有问题