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In mathematics, the Sieve of Eratosthenes (Greek: κόσκινον Ἐρατοσθένους) is a simple, ancient algorithm for finding all prime numbers up to a specified integer.[1] It works efficiently for the smaller primes (below 10 million).[2] It was created by Eratosthenes, an ancient Greek mathematician. However, none of his mathematical works survived—the sieve was described and attributed to Eratosthenes in the Introduction to Arithmetic by Nicomachus.

A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself.

[素数的定义是,一个仅能被1和本身整除的自然数]

To find all the prime numbers less than or equal to a given integer n by Eratosthenes' method:

[可以通过以下的埃拉托 {MOD}尼筛选法寻求所有小于整数n的素数]

Create a list of consecutive integers from two to n: (2, 3, 4, ..., n). [列出从2到n的一串连续自然数]
Initially, let p equal 2, the first prime number. [开始,先把2当作第一个素数,并赋值给变量p]
Strike from the list all multiples of p less than or equal to n. (2p, 3p, 4p, etc.) [将该列自然数中划掉p的倍数]
Find the first number remaining on the list after p (this number is the next prime); replace p with this number. [将剩下的自然数按原来顺序重新组成新的一列,并将第一个数赋给变量p]
Repeat steps 3 and 4 until p2 is greater than n. [重复第三第四步,直到p的平方大于n]
All the remaining numbers in the list are prime. [剩下的自然数就是所有小于n的素数]


public static List find_prime_below_number(long number){
int exe_number = 0;
boolean close = false;
List numberSet = new ArrayList();
List resultSet = new ArrayList();

for(long i=3; i numberSet.add(i);
exe_number++;
}
double half = Math.sqrt(number);
do{
Long curN = numberSet.get(0);
resultSet.add(curN);
for(int j=0; j exe_number++;
long l = numberSet.get(j);
if(l%curN==0){
numberSet.remove(j);
}
}
if(curN>half){
close = true;
}
}while(!close&&numberSet.size()!=0);

if(numberSet.size()!=0){
for(Long n : numberSet){
exe_number++;
resultSet.add(n);
}
}
System.out.println("exe_number:"+exe_number);
return resultSet;
}

当找1000000以下的素数时，执行就挂了
肯定有问题
因为原算法说了是10 million
以下的数字~
一千万以下啊~
想办法解决