1015. Reversible Primes (20)-PAT甲级真题

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
题目大意：如果一个数本身是素数，而且在d进制下反转后的数在十进制下也是素数，就输出Yes，否则就输出No

分析：判断输入是否为负数，判断n是否为素数，把n转换为d进制再反过来转换为10进制，判断是否为素数

#include <cstdio>
#include <cmath>
using namespace std;
bool isprime(int n) {
if(n <= 1) return false;
int sqr = int(sqrt(n * 1.0));
for(int i = 2; i <= sqr; i++) {
if(n % i == 0)
return false;
}
return true;
}
int main() {
int n, d;
while(scanf("%d", &n) != EOF) {
if(n < 0) break;
scanf("%d", &d);
if(isprime(n) == false) {
printf("No\n");
continue;
}
int len = 0;
int arr[100];
do{
arr[len++] = n % d;
n = n / d;
}while(n != 0);
for(int i = 0; i < len; i++) {
n = n * d + arr[i];
}
if(isprime(n) == false) {
printf("No\n");
} else {
printf("Yes\n");
}
}
return 0;
}