1015. Reversible Primes (20)

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本文介绍了一种算法，用于判断一个十进制或其它进制下的正整数是否为可逆素数，即该数及其在指定进制下的反转形式均为素数。通过将数字转换到指定进制、反转并再次转换回十进制来实现，同时验证原始及反转后的数值是否为素数。

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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

题目本身不难，但是容易理解出错。
1、输入n，若n为负数，退出，为正数，进行判断。
2、将n转化到r进制，反转，再转化到10进制，得到n2，若n和n2都为素数，输出Yes，否则输出No。

#include<iostream>
#include<string>
#include<math.h>
using namespace std;
bool isprime(int  n){
    if(n==1)return false;
    if(n==2) return true;
    for(int i=2;i<=sqrt(double(n));i++){
        if(n%i==0)
            return false;
    }
    return true;
}
int numr(int n,int r){
    int k=1,ans=0;
    while(n>0){
        ans+=(n%10)*k;
        k*=r;
        n/=10;
    }
    return ans;
}
int reverse(int n){
    int ans=0;
    while(n>0){
        ans=ans*10+n%10;
        n/=10;
    }
    return ans;
}
int reverse_r(int n,int r){
    int k=1,ans=0;
    while(n>0){
        ans+=(n%r)*k;
        k*=10;
        n/=r;
    }
    return reverse(ans);
}

int main(){
    while(1){   
        int n,r;
        cin>>n;
        if(n<0) break;
        cin>>r;
        int n1=reverse_r(n,r);
        int n2=numr(n1,r);
        if(isprime(n2)&&isprime(n))
            cout<<"Yes"<<endl;
        else cout<<"No"<<endl;
    }
    return 0;
}