PAT Grade 1015 Reversible Primes (20 minutes) (binary conversion determination and primes (since wrong forgot =))

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 (<) and D (1), 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

Subject to the effect:

Initially misunderstood the meaning of problems. . . .
In the D-ary number N is not double prime. Dual prime number is the number itself and the countdown are all prime numbers.

Implementation: determining N is prime. If not, output No, otherwise the number fell to the D-band to come over and into a decimal number, determine whether a prime number. If so, the output Yes, otherwise the output No.

复习判断素数知识点 注意 ‘ = ’ ！！！

#include <bits / STDC ++ H.>
 the using  namespace STD;
 BOOL Prime ( int X) {
     IF (X == 0 || == X . 1 ) {
         return  to false ; 
    } 
    IF (X == 2 ) {
         return  to true ; 
    } 
    for ( int I = 2 ; I <= sqrt (X); I ++) { // this place forgotten = number! ! ! 
        IF (% I X == 0 ) {
             return  to false ; 
        } 
    } 
    return  to true ; 
}
int main () 
{ 
    int A;
     int D;
     the while (CIN >> A) 
    { 
        IF (A < 0 ) {
             BREAK ; 
        } 
        CIN >> D;
         // first determination itself is not a prime number 
        IF (! Prime (A)) { 
            COUT << " No " << endl;
             Continue ; 
        } 
        // The decimal turn correspondingly 
        String S = "" ;
         int X;
         the while(A) {             
            S + = char (% d + A ' 0 ' ); 
            A = A / d; 
        } 
        // COUT S << << endl;     

        // reverse then it is converted to the binary decimal d 
        int L = s.length (); 
        X = 0 ;
         for ( int I = 0 ; I <L; I ++ ) { 
            X = X * D + S [I] - ' 0 ' ; 
        } 
        // COUT << X < <endl; 
        IF (Prime (X)) { 
            COUT << "Yes"<<endl;
        }    
        else{
            cout<<"No"<<endl;
        }
    }
    return 0;
 }