对称素数

àì夳堔傛蜴生んèń 2023-06-16 08:07 28阅读 0赞

*  **1、对称三位素数**  
    **判断三位数是否为对称素数，若是，输出Yes,否则输出No**

一、方法一
    #include<iostream>
    using namespace std;
    //对称三位素数
    
    void main(){
        
    	
    	int n;
    	cin >> n;
    	cout << "please input"<<n<<"positive inegers"<<endl;
    	int a[20];
    	for (int i = 0; i < n; i++) {
        
    		cin >> a[i];
    	}
    	for (int j = 0; j < n; j++) {
        
    		//判断a[j]是否为三位对称素数
    		for (int k = 2; k < a[j]; k++) {
        
    			if (a[j] % k == 0 || a[j] < 100 || a[j]>1000||a[j]%10!=a[j]/100%10) {
        
    				cout << "NO" << endl;
    				break;
    			}
    			if(j+1==k)
    				cout << "YES" << endl;
    		}
    	}
    	
    	
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4OTAyNjc4_size_16_color_FFFFFF_t_70]

二、方法二
    #include<iostream>
    using namespace std;
    //对称三位素数
    bool isPrime(int n) {
        
    	for (int i = 2; i < n; i++) {
        
    		if (n % i == 0)
    			return false;
    	}
    	return true;
    }
    void main(){
        
    	
    	int n;
    	while (cin >> n) {
        
    		cout << (n > 100 && n < 1000 && n / 100 % 10 == n % 10 && isPrime(n) ? "Yes\n" : "No\n");
    	}
    	
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4OTAyNjc4_size_16_color_FFFFFF_t_70 1]

*  **判断五位以内的对称素数**  
    **判断5位以内的对称素数,若是则输出Yes,否则输出No**

#include<iostream>
    using namespace std;
    //判断5位以内的对称素数,若是则输出Yes,否则输出No
    int isPrime(int n) {
        
    	int k = (int)sqrt(n);
    	if (n == 1) return 0;
    	for (int i = 2; i <= k; i++) {
        
    		if (n % i == 0)
    			return 0;
    	}
    	return 1;
    }
    void main(){
        
    	int n;
    	while (cin >> n) {
        
    		if (n / 10 == 0 && isPrime(n==1)) {
        //一位素数均为对称素数
    			cout << "Yes" << endl;
    		}
    		else if (n / 100 == 0 && (n % 10 == n / 10 % 10) && isPrime(n)==1) {
        
    			cout << "Yes" << endl;
    		}
    		else if (n / 1000 == 0 && (n % 10 == n / 100 % 10) && isPrime(n)==1) {
        
    			cout << "Yes" << endl;
    		}
    		else if (n / 10000 == 0 && (n % 10 == n / 1000 % 10 && n / 10 % 10 == n / 100 % 10) && isPrime(n)==1) {
        
    			cout << "Yes" << endl;
    		}
    		else {
        
    			cout << "No" << endl;
    		}
    	}
    	
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4OTAyNjc4_size_16_color_FFFFFF_t_70 2]

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