找质数算法（Sieve of Eratosthenes筛法）

由于一个合数总是可以分解成若干个质数的乘积，那么如果把质数（最初只知道2是质数）的倍数都去掉，那么剩下的就是质数了。
例如要查找100以内的质数，首先2是质数，把2的倍数去掉；此时3没有被去掉，可认为是质数，所以把3的倍数去掉；再到5，再到7，7之后呢，因为8，9，10刚才都被去掉了，而100以内的任意合数肯定都有一个因子小于10（100的开方），所以，去掉，2，3，5，7的倍数后剩下的都是质数了。
用程序可以这样解决，引入布尔类型数组a[i]，如果i是质数，a[i]=true，否则a[i]=false。那么划掉i可以表示成a[i]=false。
 //找出n以内质数
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());
            }
        } 

如果去掉最后一个用来显示结果的循环的话，运行Sieve(10000000)只要1秒多，而上次那个算法PrimeNum(10000000)却要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         primes[a] = null;
    }

    // Iterate from the lowest prime (2) to upperBound to flag prime or not prime.
    for (var i=2; 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                 primes[j] = false;
            }
        }
    }

    // Iterate and add up all primes.
    for (var p=0; p         if (primes[p] === true) {
            primeSum += p;
        }
    }

    return primeSum;
};

console.log(getSolutionViaSieve({upperBound: 10}));
console.log(getSolutionViaSieve({upperBound: 2000000}));