I ve got the following code inside a <script> tag on a webpage with nothing else on it. I m afraid I do not presently have it online. As you can see, it adds up all primes under two million, in two different ways, and calculates how long it took on average. The variable howOften is used to do this a number of times so you can average it out. What puzzles me is, for howOften == 1, method 2 is faster, but for howOften == 10, method 1 is. The difference is significant and holds even if you hit F5 a couple of times.

我的问题是:如何来?

(这一员额已经编辑,以纳入现成建议。) 但是,这并没有改变! 我现在非常pu。

(再一次:howOften>> 或1000多处,时间似乎稳定。) 答案似乎正确。

function methodOne(maxN) {
    var sum, primes_inv, i, j;

    sum = 0;
    primes_inv = [];
    for ( var i = 2; i < maxN; ++i ) {
        if ( primes_inv[i] == undefined ) {
            sum += i;
            for ( var j = i; j < maxN; j += i ) {
                primes_inv[j] = true;
            }
        }
    }
    return sum;
}

function methodTwo(maxN) {
    var i, j, p, sum, ps, n;

    n = ((maxN - 2) / 2);
    sum = n * (n + 2);
    ps = [];
    for(i = 1; i <= n; i++) {
        for(j = i; j <= n; j++) {
            p =  i + j + 2 * i * j;
            if(p <= n) {
                if(ps[p] == undefined) {
                    sum -= p * 2 + 1;
                    ps[p] = true;
                }
            }
            else {
                break;
            }
        }
    }
    return sum + 2;
}



// ---------- parameters
var howOften = 10;
var maxN = 10000;

console.log( iterations:  , howOften);
console.log( maxN:  , maxN);


// ---------- dry runs for warm-up
for( i = 0; i < 1000; i++ ) {
    sum = methodOne(maxN);
    sum = methodTwo(maxN);
}

// ---------- method one
var start = (new Date).getTime();

for( i = 0; i < howOften; i++ )
    sum = methodOne(maxN);

var stop = (new Date).getTime();
console.log( methodOne:  , (stop - start) / howOften);

// ---------- method two

for( i = 0; i < howOften; i++ )
    sum = methodTwo(maxN);

var stop2 = (new Date).getTime();
console.log( methodTwo:  , (stop2 - stop) / howOften);