PAT甲级1015. Reversible Primes (20)

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 (< 105) and D (1 < D <= 10), 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

Customized Test Case
Case #1:
Input:
2 2
Output:
No
(2在10进制下就是素数，转化成二进制还是素数，素数是和进制表示无关的；2的二进制表示是10，反过来是1，1不是素数)
Case #2:
Input:
32 10
Output:
No

#include <iostream>using namespace std;#include <vector>/*素数判断*/ bool isPrime(int N){    if(N==1) return false;    if(N==2) return true;    for(int i=2;i*i<=N;i++){        if(N%i==0) return false;     }    return true;}/*进制转换并Reverse*/ int Convert_to_R(int N,int R){    /*求N在R进制下的各位数字并逆序保存在vector中*/    vector<int> Ans;    while(N>0){        Ans.push_back(N%R);        N/=R;    }    /*Reverse后的R进制数转化为10进制*/    int result=0;    for(int i=0;i<Ans.size();i++) {        result=Ans[i]+result*R;    }    return result;}int main(){    int num;    cin>>num;     while(num>0){        int radix;        cin>>radix;        if(!isPrime(num)) cout<<"No"<<endl;        else {            int reverse=Convert_to_R(num,radix);            if(isPrime(reverse)) cout<<"Yes"<<endl;        else cout<<"No"<<endl;        }        cin>>num;    }    return 0;}