【PAT 甲级】1015 Reversible Primes (20)（进制转换，素数判断）

题目链接

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 (< 10^5^) 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

代码：

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#include <iostream>
#include <sstream>
#include <string>
#include <cstdio>
#include <queue>
#include <cstring>
#include <map>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e5+5;
const ll INF = 0x3f3f3f3f;
const double eps=1e-4;
const int T=3;

bool isPrime[N];
int convert(int n,int r) {
    int res=0;
    while(n) {
        res=res*r+n%r;
        n/=r;
    }
    return res;
}
void is_prime() {
    for(int i=2; i<N; i++) {
        isPrime[i]=1;
    }
    for(int i=2; i*i<=N; i++) {
        if(isPrime[i]) {
            for(int j=i*i; j<=N; j+=i) {
                isPrime[j]=0;
            }
        }
    }
}
int main() {
    int n,d;
    is_prime();
    while(cin>>n&&n>=0) {
        cin>>d;
        int res=convert(n,d);
        if(isPrime[n]&&isPrime[res]) {
            cout<<"Yes";
        } else {
            cout<<"No";
        }
        cout<<endl;
    }
    return 0;
}