PAT 1015

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<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

---

First, the item analysis:

Input decimal numbers hexadecimal N & D, if N is a prime number & D according to the number after the decimal conversion is a prime number, otherwise, outputs the output Yes No.

Second, the problem-solving ideas:

• It determines whether the prime number N (N is a prime number can perform the next operation)
• The N D from decimal to binary conversion & num1 be bitwise inverted num2
• Num2 The conversion from binary to decimal number result D
• Determining whether the result is a prime number

Points to note:

• Prime decision:

1. 1 is neither bonded nor prime number (to be determined individually, or go through the test point 2)
2. The cycling conditions for beginners for (int i = 2; i <= (int) sqrt (num1 * 1.0); i ++)

• Decimal conversion:

1. Decimal -> D Hex: In accordance with the principle of the conversion formula (do not repeat them here)
2. D Hex -> Decimal: take over the storage cycle of the output array
3. This problem not only to binary conversion, the conversion take several bit inversion: can cycle modulo modulo operation simplified procedures / memory array to facilitate the calculation

#include <iostream>
#include <math.h>
using namespace std;

int convert(int num1,int radix){
    //直接转化为对应的进制数，并实现了数位反转
    int num2=0;
    int k=0;
    int result=0;
    while(num1/radix){
        num2=num2*10+num1%radix;
        num1/=radix;
        k++;
    }
    num2=num2*10+num1;
    k++;
    //反转后的进制数转化回十进制数
    for(int i=0;i<k;i++){
        result+=(num2%10)*pow(radix,i);
        num2/=10;
    }
    return result;
}
bool isprime(int num1){ //素数的判断
    if(num1==1) //1不是素数也不是合数，要单独判断（否则测试点2过不去）
        return false;
    for(int i=2;i<=(int)sqrt(num1*1.0);i++){
        if(num1%i==0)
            return false;
    }
    return true;
}
int main()
{
    int num1; //十进制数
    int num2; //转化后的数
    int radix; //基数、进制数

    cin>>num1;
    while(num1>=0){
        cin>>radix;
        if(!isprime(num1))
            cout<<"No"<<endl;
        else {
            num2=convert(num1,radix);
            //cout<<"num2:"<<num2<<endl;
            if(isprime(num2))
                cout<<"Yes"<<endl;
            else
                cout<<"No"<<endl;
        }
        cin>>num1;
    }
    return 0;
}