C程序在数组中找到最小和最大的质数

问题陈述

给定一个包含 n 个正整数的数组。我们必须找到素数 具有最小值和最大值的数字。

如果给定的数组是 -

arr [] = {10, 4, 1, 12, 13, 7, 6, 2, 27, 33}
then minimum prime number is 2 and maximum prime number is 13

算法

1. Find maximum number from given number. Let us call it maxNumber
2. Generate prime numbers from 1 to maxNumber and store them in a dynamic array
3. Iterate input array and use dynamic array to find prime number with minimum and maximum value

Example

的中文翻译为：

示例

#include <iostream>
#include <vector>
#include <climit>
#define SIZE(arr) (sizeof(arr) / sizeof(arr[0]))
using namespace std;
void printMinAndMaxPrimes(int *arr, int n){
   int maxNumber = *max_element(arr, arr + n);
   vector<bool> primes(maxNumber + 1, true);
   primes[0] = primes[1] = false;
   for (int p = 2; p * p <= maxNumber; ++i) {
      if (primes[p]) {
         for (int i = p * 2; i <= maxNumber; i += p) {
            primes[p] = false;
         }
      }
   }
   int minPrime = INT_MAX;
   int maxPrime = INT_MIN;
   for (int i = 0; i < n; ++i) {
      if (primes[arr[i]]) {
         minPrime = min(minPrime, arr[i]);
         maxPrime = max(maxPrime, arr[i]);
      }
   }
   cout << "Prime number of min value = " << minPrime << "</p><p>";
   cout << "Prime number of max value = " << maxPrime << "</p><p>";
}
int main(){
   int arr [] = {10, 4, 1, 12, 13, 7, 6, 2, 27, 33};
   printMinAndMaxPrimes(arr, SIZE(arr));
   return 0;
}

输出

当您编译并执行上述程序时，它会生成以下输出 −

Prime number of min value = 2
Prime number of max value = 13

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