PAT甲级真题 1015 Reversible Primes (20分) C++实现

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本文介绍了一种算法，用于检测一个数在特定进制下是否为可逆素数，即该数及其在该进制下的反转数均为素数。通过将十进制数转换为目标进制，并利用vector保存反转后的数，再判断其是否为素数。

题目

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

将10进制数转为相应进制的数字，顺序保存到vector中，再利用n = n * radix + v[i]求出倒置后的10进制数值，判断其是否是素数即可。

测试点1考察的是：1不是素数。

代码

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;

bool isPrime(int n){
    bool ret = true;
    if (n==1){ 
        ret = false;
    }
    for (int i=2; i<=sqrt(n); i++){
        if (n%i==0){
            ret = false;
            break;
        }
    }
    return ret;
}

int main(){
    int n, d;
    cin >> n;
    while (n > 0){
        cin >> d;
        if (!isPrime(n)){
            cout << "No" << endl;
        }
        else {
            vector<int> a(32);
            int len = 0;
            while (n!=0){
                a[len++] = n % d;
                n /= d;
            }
            for (int i=0; i<len; i++){
                n = n * d + a[i];
            }
            if (isPrime(n)){
                cout << "Yes" << endl;
            }
            else {
                cout << "No" << endl;
            }
        }
        cin >> n;
    }
    return 0;
}