半质数的个数

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

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

	bool * bPrime = (bool *)malloc(y * sizeof(bool));
	bPrime[0] = false;
	bPrime[1] = false;
	for(int i = 2; i < y; i++)
	{
		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)
			{
				bPrime[j] = false;
			}
		}
	}

	int nPrime = 0;
	for(int i = 2; i < y; i++)
	{
		if(bPrime[i])
			nPrime++;
	}


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

	int nHalfPrime = 0;
	for(int i = 0; i < nPrime; i++)
	{
		if(y / Prime[i] < Prime[i])
		{
			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()
{    
	printf("%d\n",getNum(2, 2000000));
}

质数是大家熟知的概念,我们定义一个半质数的概念:如果一个数恰好是两个质数的乘积(可以相同),则称它为半质数。前几个半质数是 4, 6, 9, 10, 14, 15, 21, 22, 25, 26。我们的问题是,输入两个正整数x<=y,问[x,y]之间有多少个半质数? 输入:x,y 输出:[x,y]之间有多少个半质数。 输入数据范围 1<=x<=y<=2000000。 


代码:


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转载自blog.csdn.net/Yshe_xun/article/details/19337889