使用递归计算并打印前 N 个质数

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英文:

How to calculate and print first N Prime Numbers Using Recursion

问题

我写了以下递归代码,但结果很奇怪。请帮忙找出我的 C 代码中的错误。

例如,如果我将 n 的值设为 10,那么输出应该是 2,3,5,7,11,13,17,19,23,29,但是我的代码输出是 2 3 5 7 9 11 13 15 17 19

啊!我尝试了很长时间,但找不到问题。这是我的递归 C 代码:

void primeNumbers (int n)
{
    if (n == 0)  
        return;

    static int i = 2;
    
    if (isPrime(i))      
    {
        printf("%d ", i);
        i++;                
        primeNumbers(n - 1);
    }
    else 
    {
        i++;
        primeNumbers(n);
    }
}

检查数字是否为素数的函数:

bool isPrime(int n)
{
    // 边缘情况,不是基本情况
    if (n <= 1)
        return false;

    static int i = 2; 
    
    if (i > n / 2) 
        return true;
    
    if (n % i == 0) 
        return false;
    
    i++;
    return isPrime(n); // 用 i+1 的值再次调用
}

输出:[如果 N 为 10]

2 3 5 7 9 11 13 15 17 19

如果我将 n 的值设为 10,那么输出应该是 2,3,5,7,11,13,17,19,23,29,但是我的代码输出是 2 3 5 7 9 11 13 15 17 19

英文:

I wrote the following recursive code but the outcome is strange. Please Help to find the error in my C code.

For example if I take the value of n as 10 then the output should be 2,3,5,7,11,13,17,19,23,29
but with to my code the output is 2 3 5 7 9 11 13 15 17 19

Ahh! I tried for a long time but couldn't find the problem. Here is my recursive C code:

void primeNumbers (int n)
{
    if (n == 0)  
        return;

    static int i = 2;
    
    if (isPrime(i))      
    {
        printf(&quot;%d &quot;, i);
        i++;                
        primeNumbers(n - 1);
    }
    else 
    {
        i++;
        primeNumbers(n);
    }
}

The function to check Number is Prime or not:

bool isPrime(int n)
{
    // edge case, not a base case
    if (n &lt;= 1)
        return false;

    static int i = 2; 
    
    if (i &gt; n / 2) 
        return true;
    
    if (n % i == 0) 
        return false;
    
    i++;
    return isPrime(n); // call again with i+1 value
}

Output: [if N is 10]

2 3 5 7 9 11 13 15 17 19

if I take the value of n as 10 then the output should be 2,3,5,7,11,13,17,19,23,29 but with to my code the output is 2 3 5 7 9 11 13 15 17 19.

答案1

得分: 2

问题在于您使用一个永远不会重置为2static变量i作为除数,用于测试下一个数字。

以下是isPrime的另一种递归实现:

bool testPrime(int n, int div) {
    if (n / div < div)
        return true;
    else if (n % div == 0)
        return false;
    else
        return testPrime(n, div + 1);
}

bool isPrime(int n) {
    return n >= 2 && testPrime(n, 2);
}

递归函数可以简化为单个表达式:

bool testPrime(int n, int div) {
    return (n / div < div) || (n % div != 0 && testPrime(n, div + 1));
}

由于相同的原因,枚举函数也是不正确的:static变量i从未重置为2,所以该函数只会工作一次,而且在测试和打印i后,您以n - 1的递归方式调用它,因此它会在中途停止。

对于primeNumbers(),可以采用相同的方法将循环转换为递归调用:

void printPrimes(int p, int n) {
    if (n > 0) {
        if (isPrime(p)) {
            printf("%d ", p);
            printPrimes(p + 1, n - 1);
        } else {
            printPrimes(p + 1, n);
        }
    }
}

void primeNumbers(int n) {
    printPrimes(2, n);
    printf("\n");
}
英文:

The problem is you use a static variable i for the divisor that never gets reset to 2 for the next number to test.

Here is an alternative recursive implementation of isPrime:

bool testPrime(int n, int div) {
    if (n / div &lt; div)
        return true;
    else
    if (n % div == 0)
        return false;
    else
        return testPrime(n, div + 1);
}

bool isPrime(int n) {
    return n &gt;= 2 &amp;&amp; testPrime(n, 2);
}

The recursive function can be simplified as a single expression:

bool testPrime(int n, int div) {
    return (n / div &lt; div) || (n % div != 0 &amp;&amp; testPrime(n, div + 1));
}

The enumerating function is also incorrect for the same reason: static variable i is never reset to 2 so the function would only work once, and you call it recursively with n - 1 after testing and printing i so it stops enumerating half way.

The same approach can be taken for primeNumbers() to convert a loop into recursive calls:

void printPrimes(int p, int n) {
    if (n &gt; 0) {
        if (isPrime(p)) {
            printf(&quot;%d &quot;, p);
            printPrimes(p + 1, n - 1);
        } else {
            printPrimes(p + 1, n);
        }
    }
}

void primeNumbers(int n) {
    printPrimes(2, n);
    printf(&quot;\n&quot;);
}

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  • 本文由 发表于 2023年7月10日 15:11:09
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