//求素数,该方法是利用大于6的素数可以表示为6n+1或6n+5,而它的约数应为小于等于它的平方根
#include<stdio.h>
#include<math.h>
const int Max = 1000;
void Method_1();
void Method_2();
void main()
{
Method_1();
// Method_2();
}
void Method_1()
{
int i = 1;
int j = 1;
int flag = 1;
int Array[300] = {2,3,5};
int count = 3;
for( i = 7; i < Max; )
{
int temp = (int)sqrt(i);
for (j = 0; Array[j]<=temp; j++ )
//小于等于i的平方根的素数,本应有j<count,但n与2n间一定有一个素数,n*n大于2n,中间一定有一个素数,故不会越界
{
if(i%Array[j]==0)
{
flag = 0;//查找的到约数立即停止
break;
}
}
if(flag==1)
{
Array[count] = i;
count++;
}
flag = 1;
i += 4; //6n+1与6n+5才可能是素数
temp = (int)sqrt(i);
for (j = 0;Array[j]<=temp; j++ )
{
if(i%Array[j]==0)
{
flag = 0;
break;
}
}
if(flag==1)
{
Array[count] = i;
count++;
}
flag = 1;
i+=2;
}
for( i = 0; i < count; i++ )
{
printf("%4d ",Array[i]);
if(i%10==9) printf("\n");
}
printf("\n");
}
//埃拉托斯特尼筛法
void Method_2()
{
int A[1001]={0};
int temp = 0;
int i = 0;
int j = 0;
int count = 1;
A[1] = 1;
A[0] = 1;
for( i = 2; i < 1001; i++ )
A[i] = 0;
for ( i = 2; i <= 500; i++)
{
int temp = 1000/i;
for (j = 2;j<= temp;j++)
{
A[i*j] = 1;
}
}
for( i = 0; i < 1001;i++)
{
if(A[i]==0)
{
printf("%4d ",i);
count++;
if(count%10==1) printf("\n");
}
}
printf("\n");
}