A1015. Reversible Primes

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

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
int numReverse(int N, int radix){
    int num[50], temp, index = 0, ans = 0;
    do{
        temp = N % radix;
        num[index++] = temp;
        N = N / radix;
    }while(N != 0);
    for(int i = index - 1, P = 1; i >= 0; i--){
        ans += num[i] * P;
        P = P * radix;
    }
    return ans;
}
int isPrime(int N){
    int sqr = (int)sqrt(N * 1.0);
    if(N == 1)
        return 0;
    for(int i = 2; i <= sqr; i++){
        if(N % i == 0)
            return 0;
    }
    return 1;
}
int main(){
    int N, D, N2;
    while(1){
        scanf("%d", &N);
        if(N < 0)
            break;
        scanf("%d", &D);
        int temp = numReverse(N, D);
        if(isPrime(N) && isPrime(temp))
            printf("Yes
");
        else printf("No
");
    }
    cin >> N;
    return 0;
}

总结：

1、判断素数：

int isPrime(int N){
    int sqr = (int)sqrt(N * 1.0);  //N应该转换为小数
    if(N == 1)        //当N = 1时应返回false，容易忽略，1不是素数
        return 0;
    for(int i = 2; i <= sqr; i++){   //i <= sqr； i从2开始查找
        if(N % i == 0)
            return 0;
    }
    return 1;
}