C程序有多少个素数

Question

我写了这个程序来查找 1 到 50000 之间的素数，但我仍然需要找出有多少个素数（我尝试了很多技巧，但我没有成功）

#include <stdio.h>

//int getValueFromUser();
void PrintListOfPrime(int value);

int main() {
    int value = 23;
    PrintListOfPrime(value);
    return 0;
}

void PrintListOfPrime(int value) {
    int  ValueIsPrime; //ValueIsPrime is used as flag variable

    printf("The list of primes: ");

    for (int i = 2; i <= value; i++) {
        ValueIsPrime = 1;
        /* Check if the current number i is prime or not */
        for (int j = 2; j <= i / 2; j++) {
            /*
             * If the number is divisible by any number
             * other than 1 and self then it is not prime
             */

            if (i % j == 0) {
                ValueIsPrime = 0;
                break;
            }
        }

        /* If the number is prime then print */
        if (ValueIsPrime == 1)
            printf("%d, ", i);
    }
    printf("\n");
}

Answer 1

我尝试了很多技巧，但我没有成功

如果 OP 的代码运行时间过长，则迭代到 i 的平方根，而不是 i/2。

j <= i / 2 很慢。请改用j <= i / j。

---

对每个素数形成一个计数和增量。 @gspr

if (ValueIsPrime == 1) {
  printf("%d, ", i);
  prime_count++;
}

---

“找到 1 到 50000 之间的素数”的更大变化更快，研究Sieve of Eratosthenes

Answer 2

你好，快速的答案是在main中创建一个变量，int totaleOfPrimes = 0;例如。 然后通过引用函数发送它：

1. 函数声明：void PrintListOfPrime(int value,int* counter);
2. 函数调用：void PrintListOfPrime(value,&totaleOfPrimes); 然后在打印前增加计数器：

if (ValueIsPrime == 1){
   (*counter)++;
   printf("%d, ", i);
}

Answer 3

没有必要为 2 和 value 之间的所有数字迭代循环。您应该只考虑 2 和奇数。

该函数可以如下面的演示程序所示。

#include <stdio.h>

static inline size_t PrintListOfPrime( unsigned int n ) 
{
    size_t count = 0;

    printf( "The list of primes:\n" );

    for ( unsigned int i = 2; i <= n; i = i != 2 ? i + 2 : i + 1 ) 
    {
        int isPrime = 1;
        /* Check if the current number i is prime or not */
        for ( unsigned int j = 3; isPrime && j <= i / j; j += 2 ) 
        {
            /*
             * If the number is divisible by any number
             * other than 1 and self then it is not prime
             */

            isPrime = i % j != 0;
        }

        /* If the number is prime then print */
        if ( isPrime )
        {
            if ( ++count % 14 == 0 ) putchar( '\n' ); 
            printf( "%u ", i );
        }
    }

    return count;
}

int main(void) 
{
    unsigned int n = 50000;

    size_t count = PrintListOfPrime( n );
    
    printf( "\n\nThere are %zu prime numbers up to %u\n", count, n );
    
    return 0;
}

Answer 4

在 C 中运行此代码。它将返回 pi(x) 函数的值。基本上就是 Prime 计数功能：

#include <stdio.h>
#define LEAST_PRIME 2
#include <math.h>

int main() //works for first 10000 primes.
{
int lower_limit = 2, no_of_sets;
// printf("NUMBER OF SETS: ");
// scanf("%d", &no_of_sets);

int remainder, divisor = 2, remainder_dump, upper_limit; //upper limit to be specified 
                                                         //by user.
int i = 1;
// printf("SPECIFY LOWER LIMIT: ");
// scanf("%d", &lower_limit);
int number_to_be_checked = lower_limit;
printf("SPECIFY UPPER LIMIT: ");
scanf("%d", &upper_limit);

printf("2\t\t\t\t", number_to_be_checked);
//PRINTS 2.*/
do
{
    remainder_dump = 1;
    divisor = 2;

    do
    {
        remainder = number_to_be_checked % divisor;
        if (remainder == 0)
        {
            remainder_dump = remainder_dump * remainder; // dumping 0 for rejection.
            break;
        }
        ++divisor;

    } while (divisor <= number_to_be_checked / divisor); // upto here we know number 
is prime or not.
    if (remainder_dump != 0)
    {
        ++i;
        printf("%d.\t\t\t\t", number_to_be_checked); //print if prime.
    };
    number_to_be_checked = number_to_be_checked + 1;
} while (number_to_be_checked <= upper_limit);
printf("\n                                    pi(x) = %d \n", i);
//printf("pi function value is %f.", (i - 1) / (log(i - 1)));
float app;
app = upper_limit / (log(upper_limit));
float plot_value;
plot_value = (i) / app;
printf("                                    BETA FUNCTION VALUE ~ %f", plot_value);
return 0;
}