找質數演算法（Sieve of Eratosthenes篩法）
例如要查詢100以內的質數，首先2是質數，把2的倍數去掉；此時3沒有被去掉，可認為是質數，所以把3的倍數去掉；再到5，再到7，7之後呢，因為8，9，10剛才都被去掉了，而100以內的任意合數肯定都有一個因子小於10（100的開方），所以，去掉，2，3，5，7的倍數後剩下的都是質數了。
用程式可以這樣解決，引入布林型別陣列a[i]，如果i是質數，a[i]=true，否則a[i]=false。那麼劃掉i可以表示成a[i]=false。
//找出n以內質數
void Sieve(int n)
{
bool[] a = new bool[n+1];
for (int i = 2; i <= n; i++) a[i] = true;
for (int i = 2; i <= Math.Sqrt(n); i++)
{
if (a[i])
for (int j = i; j*i <= n; j++) a[j * i] = false;
}
for (int i = 0; i <= n; i++)
{
if (a[i])
Console.Write("{0},",i.ToString());
}
}
如果去掉最後一個用來顯示結果的迴圈的話，執行Sieve(10000000)只要1秒多，而上次那個演算法PrimeNum(10000000)卻要71秒多！
/**
* Solves the challenge by utilizing a Sieve of Eratosthenes (https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes)
*
* @param options
* @returns {number}
*/
getSolutionViaSieve = function (options) {
var upperBound = options.upperBound || 0,
primes = new Array(upperBound),
primeSum = 0;
// Initialize all of the numbers from zero to upper bound with a null value. Later, each index will be checked
// for null to see if it's been market yet.
for (var a=0; a<upperBound; a++) {
primes[a] = null;
}
// Iterate from the lowest prime (2) to upperBound to flag prime or not prime.
for (var i=2; i<upperBound; i++) {
// Only proceed if the current index hasn't been flagged as prime or not prime.
if (primes[i] === null) {
// Flag the current index as prime.
primes[i] = true;
// And flag any remaining multiples of this prime as not prime.
for (var j=i*i; j<upperBound; j+=i) {
primes[j] = false;
}
}
}
// Iterate and add up all primes.
for (var p=0; p<upperBound; p++) {
if (primes[p] === true) {
primeSum += p;
}
}
return primeSum;
};
console.log(getSolutionViaSieve({upperBound: 10}));
console.log(getSolutionViaSieve({upperBound: 2000000}));