C编程为数组中的最小和最大素数
问题陈述
给定n个正整数数组。我们必须找到具有最小值和最大值的质数。
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
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 << "\n";
cout << "Prime number of max value = " << maxPrime << "\n";
}
int main(){
int arr [] = {10, 4, 1, 12, 13, 7, 6, 2, 27, 33};
printMinAndMaxPrimes(arr, SIZE(arr));
return 0;
}输出结果
当您编译并执行上述程序时。它生成以下输出-
Prime number of min value = 2 Prime number of max value = 13