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

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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。

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#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;}