POJ3292 Semi-prime H-numbers 素數篩選

        題目大意：H-Number被定義爲4*n+1的數，其中n是整數，將H-Number分爲unit,H-Prime ,H-Composites，1是唯一的unit,H-Prime是它只能被唯一的兩個H-Number整除

分爲是1和它本身，剩下的就是H-Composites。給定一個H-Number  h，計算出1-h中有多少個H-Prime。

        仿造埃氏篩法進行H-Prime篩選就可以了。

#ifndef HEAD
#include <stdio.h>
#include <vector>
#include <math.h>
#include <string.h>
#include <string>
#include <iostream>
#include <queue>
#include <list>
#include <algorithm>
#include <stack>
#include <map>

using namespace std;
#endif // !HEAD

bool isprime[1000002];

int icount[1000002];
int main()
{
	
	memset(isprime, 1, sizeof(isprime));

	memset(icount, 0, sizeof(icount));
	isprime[1] = 0;
	for (int i = 5; i <= 1000001;i += 4)
	{
		if (isprime[i])
		{
			for (int j = i * 2; j <= 1000001;j+= i)
			{
				isprime[j] = 0;
			}
		}
	}
	vector<int> h_primes;
	h_primes.resize(89070);
	int index = 0;
	for (int i = 5; i <= 1000001; i += 4)
	{
		if (isprime[i])
		{
			//h_primes.push_back(i);
			h_primes[index++] = i;
		}
	}
	
	memset(isprime, 0, sizeof(isprime));
	for (int i = 0; i < h_primes.size();i++)
	{
		for (int j = i; j < h_primes.size();j++)
		{
			long long res = ((long long)h_primes[i]) * h_primes[j];
			if (res <= 1000001)
			{
				isprime[res] = 1;
			}
			else
				break;
		}
	}
	int count = 0;
	for (int i = 0; i <= 1000001;i++)
	{
		if (isprime[i])
		{
			count++;
		}
		icount[i] = count;
	}
#ifdef _DEBUG
	freopen("d:\\in.txt", "r", stdin);
#endif
	int a;
	while (scanf("%d\n", &a) != EOF)
	{
		if (a == 0)
		{
			break;
		}
		printf("%d %d\n", a, icount[a]);
	}
	return 0;
}