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 maxnumber2. generate prime numbers from 1 to maxnumber and store them in a dynamic array3. 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 << "
"; cout << "prime number of max value = " << maxprime << "
";}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 = 2prime number of max value = 13
以上就是c程序在数组中找到最小和最大的质数的详细内容。