半质数的个数

#include <iostream>
#include <time.h>
#include <windows.h>
using namespace std;

int Prime(int x,int y)
{
	if(x <2 || y < 2)
		return 0;
	int nCount = 0;
	for(int n=x; n<=y; ++n)
	{
		for(int i=2; i<n; ++i)
		{
			int tmp1 = n/i;
			if(tmp1 < i)
				break;
			if(tmp1 * i == n)	//如果能整出
			{
				int j = 2;
				for(j; j<tmp1; ++j)
				{
					if((tmp1%j==0) || (j<i && i%j ==0))
					{
						break;
					}
				}
				if(j>=tmp1)
				{
					++nCount;
				}
				break;
			}
		}
	}
	return nCount;
}

int main()
{
	int nMin = 2,nMax = 20000;
	printf("%d到%d的半质数:
",nMin,nMax);
	int nStar = GetTickCount(); 
	cout<<Prime(nMin,nMax)<<endl;
	int nEnd = GetTickCount();
	cout<<"time = "<<nEnd-nStar<<endl;
	system("pause");
	return 0;
} 

 

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <Windows.h>
using namespace std;

int getNum(int x, int y)
{
	if(x < 1 || x > y)
		return 0;

	bool * bPrime = (bool *)malloc(y * sizeof(bool));
	bPrime[0] = false;			// 2
	bPrime[1] = false;			// 3
	for(int i = 2; i < y; i++)  // {false,false true,true......}
	{
		bPrime[i] = true;
	}

	for(int i = 2; i < y; i ++) 
	{
		if(bPrime[i])
		{
			if(y/i < i)
			{
				break;
			}

			for(int j = i * i; j < y; j += i) // 合数
			{
				bPrime[j] = false;
			}
		}
	}

	int nPrime = 0;
	for(int i = 2; i < y; i++)		
	{
		if(bPrime[i])
			nPrime++;		//统计还剩余多少个质数
	}

	int * Prime = (int *)malloc(nPrime * sizeof(int));
	for(int i = 2, j = 0; i < y; i++)		
	{
		if(bPrime[i])
		{
			Prime[j++] = i;	//得到质数
		}
	}

	int nHalfPrime = 0;
	for(int i = 0; i < nPrime; i++)
	{
		if(y / Prime[i] < Prime[i])	//排除较大 * 较小的情况
		{
			break;
		}

		for(int j = i; j < nPrime; j++)
		{

			if(j != i && y / Prime[i] < Prime[j])
			{
				break;
			}

			int HalfPrime = Prime[i] * Prime[j];
			if(HalfPrime >= x) 
			{
				nHalfPrime++;
			}
		}
	}

	free(Prime);
	free(bPrime);
	return nHalfPrime;
}

int main()
{   
	int nStat = GetTickCount();
	printf("%d
",getNum(2, 2000000));
	printf("%d
",GetTickCount()-nStat);
	system("pause");
}