欧拉项目 #10，java，适用于小数

Question

*免责声明，当我说 "I have verified this is the correct result" 时，请解释为我已经根据 WolframAlpha 的答案检查了我的解决方案，我认为这非常准确。

*目标，求出所有小于等于2,000,000（两百万）的质数之和

*问题，只要我的测试值范围大约小于或等于

，我的代码就会输出正确的结果

一旦测试输入大于大约 1,300,000，我就不会输出正确的结果；我的输出将关闭...

测试输入：----199,999 测试输出：---1,709,600,813 正确结果：1,709,600,813

测试输入：----799,999 测试输出：---24,465,663,438 正确结果：24,465,663,438

测试输入：----1,249,999 测试输出：---57,759,511,224 正确结果：57,759,511,224

测试输入：----1,499,999 测试输出：--- 82,075,943,263 正确结果：82,074,443,256

测试输入：----1,999,999 测试输出：--- 142,915,828,925 正确结果：142,913,828,925

测试输入：----49,999,999 测试输出：--- 72,619,598,630,294 正确结果：72,619,548,630,277

*我的代码，这是怎么回事，为什么它适用于较小的输入？我什至使用 long，而不是 int...

long n = 3;
long i = 2;
long prime = 0;
long sum = 0;
while (n <= 1999999) {
  while (i <= Math.sqrt(n)) {    // since a number can only be divisible by all
                            // numbers
                            // less than or equal to its square roots, we only
                            // check from i up through n's square root!
    if (n % i != 0) {       // saves computation time
      i += 2;               // if there's a remainder, increment i and check again
    } else {
      i = 3;                // i doesn't need to go back to 2, because n+=2 means we'll
                            // only ever be checking odd numbers
      n += 2;               // makes it so we only check odd numbers
    }
  }                         // if there's not a remainder before i = n (meaning all numbers from 0
                            // to n were relatively prime) then move on
  prime = n;                // set the current prime to what that number n was
  sum = sum + prime;
  i = 3;                    // re-initialize i to 3
  n += 2;                   // increment n by 2 so that we can check the next odd number

}
System.out.println(sum+2); // adding 2 because we skip it at beginning

请帮助:)

Answer 1

问题是您没有正确检查要添加到总和中的最新素数是否小于限制。你有两个嵌套循环，但你只检查外循环的限制：

while (n <= 1999999) {

但是你没有检查内循环：

 while (i <= Math.sqrt(n)) {

然而，您在该循环内重复前进到下一个候选素数 (n += 2;)。这允许候选素数超过限制，因为在外循环的每次迭代中只检查第一个候选素数的限制，而不检查内循环访问的任何后续候选素数。

举个例子，在极限值为1,999,999的情况下，这使得the next prime after 1,999,999, which is 2,000,003. You'll note that the correct value, 142,913,828,922,比你的结果142,915,828,925少2,000,003。

更简单的结构

下面是代码结构的一种方式，使用标签和 continue 使用该标签来简化结构：

public static final long primeSum(final long maximum) {
    if (maximum < 2L) return 0L;
    long sum = 2L;

    // Put a label here so that we can skip to the next outer loop iteration.
    outerloop:
    for (long possPrime = 3L; possPrime <= maximum; possPrime += 2L) {
        for (long possDivisor = 3L; possDivisor*possDivisor <= possPrime; possDivisor += 2L) {
            // If we find a divisor, continue with the next outer loop iteration.
            if (possPrime % possDivisor == 0L) continue outerloop;
        }
        // This possible prime passed all tests, so it's an actual prime.
        sum += possPrime;
    }

    return sum;
}

欧拉项目 #10，java，适用于小数

project euler #10, java, correct for small numbers

java

primes

更简单的结构