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 (<10^5) and D (1

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 Output:

Yes
Yes
No

Sample Input:

73 10
23 2
23 10
-2

思路

思路不难。没学过C++，想用string练练手，熟悉一下STL。

如果输入的进制是10进制，则将N转为字符串后直接用reverse()进行倒置，然后判断是否为素数；

如果输入的进制是其他进制，将N化为对应进制的数并存到字符串中，然后reverse()倒置，倒置完了后再转为10进制真值，进而判断是否为素数。

代码

#include 
#include 
#include 
#include 
#include 
using namespace std;
int sushu(int num){
     
    if(num == 1)
        return 0;
    for(int i=2; i<=sqrt(num); i++){
     
        if(num % i == 0)
            return 0; 
    }
    return 1;
}
int reverse_check(int num, int radix){
     
    int real_num_reverse=0;
    string real_num_reverse_str;
    stringstream ss;
    int yushu;
    if(radix == 10){
     
        ss << num; ss >> real_num_reverse_str;
        reverse(real_num_reverse_str.begin(), real_num_reverse_str.end());
        ss.clear();
        ss << real_num_reverse_str; ss >> real_num_reverse;
        if(sushu(real_num_reverse) == 1)
            return 1;
        else
            return 0;
    }
    while(num > 0){
     
        yushu = num % radix;
        real_num_reverse_str.push_back((yushu + '0'));
        num /= radix;
    }
    real_num_reverse = 0;
    for(int i=real_num_reverse_str.size()-1; i>=0; i--){
     
        real_num_reverse += (real_num_reverse_str[i]-'0')*pow(radix, real_num_reverse_str.size()-1-i);
    }
    if(sushu(real_num_reverse) == 1)
        return 1;
    else
        return 0;
}

int main(){
     
    string str;
    int N, D;
    while(1){
     
        cin>>N;
        if(N<0) break;
        cin>>D;
        if(sushu(N) == 0){
     
            cout<<"No"<<endl;
            continue;
        }
        if(reverse_check(N, D) == 1)
            cout<<"Yes"<<endl;
        else 
            cout<<"No"<<endl;
    }
    return 0;
}