PAT 甲级 1015 Reversible Primes (20 分) （进制转换和素数判断(错因为忘了=)）

Posted 2022-06-07 caiyishuai

1015 Reversible Primes (20 分)

 

A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (<) and D (1), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line Yes if N is a reversible prime with radix D, or No if not.

Sample Input:

73 10
23 2
23 10
-2

Sample Output:

Yes
Yes
No

题目大意：

一开始理解错了题意。。。。
数字N在D进制下是不是双重素数。双重素数是本身和倒数皆为素数的数。

实现：判断N是否为素数。如果不是，输出No，否则将该数在D进制下倒过来再化为十进制数，判断是否为素数。如果是，输出Yes，否则输出No.

复习判断素数知识点 注意 ‘ = ’ ！！！

#include<bits/stdc++.h>
using namespace std;
bool prime(int x)
    if(x==0||x==1)
        return false;
     
    if(x==2)
        return true;
    
    for(int i=2;i<=sqrt(x);i++)//这个地方忘记了=号！！！ 
        if(x%i==0)
            return false;
        
     
    return true;

int main()

    int a;
    int d;
    while(cin>>a)
    
        if(a<0)
            break;
        
        cin>>d;
        //先判断本身是不是素数 
        if(!prime(a))
            cout<<"No"<<endl;
            continue;
        
        //根据相应地进制转 
        string s="";
        int x;
        while(a)            
            s+=char(a%d+‘0‘);
            a=a/d;
         
        //cout<<s<<endl;    

        //反向再把它从d进制转成10进制 
        int l = s.length();
        x=0;
        for(int i=0;i<l;i++)
            x=x*d+s[i]-‘0‘;
        
        //cout<<x<<endl; 
        if(prime(x))
            cout<<"Yes"<<endl;
            
        else
            cout<<"No"<<endl;
        
    
    return 0;