1015. Reversible Primes (20)-PAT甲级真题

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, 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];
        printf("%s", isprime(n) ? "Yes\n" : "No\n");
    }
    return 0;
}

#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, 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];

printf("%s", isprime(n) ? "Yes\n" : "No\n");

}

return 0;

}

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