PAT (Advanced Level) Practise 1015

PAT (Advanced Level) Practise 1031

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

思路

题目的意思是给出一个数N,和进制radix,首先判断N是不是素数,如果是,在将其转换为radix进制,并取逆序得到rev_num,再判断rev_num是不是素数,如果是,则为reversible prime,否则不是.

编程实现

#include <stdio.h>
#include <math.h>

typedef enum {false, true} bool;

bool prime(int num)
{
    int i, upper;
    bool is_prime = true;

    if ( num < 2 ) 
        return false;
    upper = (int)sqrt((double)num)+1;
    /* 注意i<=upper会导致num==2时,出现错误 */
    for ( i = 2; i < upper; i++ ) {
        if ( num % i == 0 ) {
            is_prime = false;
            break;
        }
    }
    return is_prime;
}

int reverse(int key, int radix)
{
    int revs = 0;

    while ( key ) {
        revs = revs*radix + (key % radix);
        key /= radix;
    }
    return revs;
}

int main(void)
{
    int num, radix;

    while ( true ) {
        scanf("%d", &num);
        if ( num < 0 ) break;
        scanf("%d", &radix);
        if ( (!prime(num)) || 
             (!prime(reverse(num, radix))) )
            printf("No\n");
        else
            printf("Yes\n");
    }

    return 0;
}