并行运行时不像我预期

Question

我正在尝试使用OpenMP实现RSA算法的并行版本。该程序将公共密钥和N作为用户输入，计算到质数并生成公共密钥和私钥对，还生成了加密和解密的消息。我瘫痪了主要功能中的公共密钥生成循环。现在，问题在于我的程序无法计算经过的时间。因此，我对我的并行实施不佳。

代码块

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<omp.h>
#include<stdbool.h>

int GCD(int , int);
long int Decryption__k(long int , int );
unsigned Decryption(unsigned , unsigned, unsigned);


int main(int argc, char* argv[])
{
    unsigned public_key, n, totient, private_key, encryption, my_msg, plainText;
    unsigned f_prime, s_prime, temp = 0;
    double start, end;
    int thread_count=strtol(argv[1], NULL, 10);
     printf("Enter value for public key: ");
    scanf("%d", &public_key);

    printf("Enter value for N: ");
    scanf("%d", &n);

    printf("\nEnter encrypted message: ");
    scanf("%u", &my_msg);

    if(my_msg >= n){
        printf("\nError enter message small than N\n");
        exit(0);
    }

    
     start = omp_get_wtime();

    for (int i = 1; i <= n; ++i) {
        if (n % i == 0) {
            if(i % 2 == 1 && i != 1 && i != n){
                f_prime = i;
                if(f_prime != temp){
                    temp = f_prime;
                    f_prime = s_prime;
                    s_prime = temp;
                }
            }
        }
    }

    totient = (f_prime - 1) * (s_prime - 1);
   
    bool flag = false;
    #pragma omp parallel num_threads(thread_count)
    #pragma omp parallel for
    for(public_key=0; public_key < totient; public_key++){
        
        if(flag)continue;
        if(GCD(public_key, totient) == 1){
            encryption = public_key;
            flag=true;
        }
    }

    private_key = Decryption__k(encryption, totient);

    printf("\nPublic Key: {%u, %d}\nPrivate Key: {%u, %d}\n Prime 1 : %d\nPrime 2: %d\n", encryption, n, private_key, n, f_prime, s_prime);


    plainText = Decryption(my_msg, private_key, n);

    printf("\nEncrypted message: %u\nDecrypted message: %u\n", my_msg, plainText);

   
   end=omp_get_wtime();
    printf("\nElapsed time =%e seconds\n", end - start);

    return 0;
}

    long int Decryption__k(long int e, int totient)
{
  int d = 1;

  while ((e * d) % totient != 1) {
    d++;
  }
  return d;
}

unsigned Decryption(unsigned message, unsigned key, unsigned n)
{
    unsigned r = 1;

    for (unsigned a = 1, p = message; a <= key; a <<= 1, p = p * p % n){
        if (key & a){
            r = r * p % n;
        }
    }

    return r;
}

int GCD(int a, int b)
{
    int temp;
    while (b != 0)
    {
        temp = a % b;

        a = b;
        b = temp;
    }
    return a;
}

输出：

根据输出的时间为NAN，

我在这里不包括GCD，解密__k和解密功能，因为我没有在那里实现催化作用。

谢谢