PAT1015:Reversible Primes

1015. Reversible Primes (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

1.先判断十进制数N是否是素数，不是直接输出No。
2.如果N是素数先把N转为D进制数N1，然后得到的D进制数N1按颠倒的位数转换成十进制数N2，检查N2是否是素数即可。
例：
  1）比如输入N = 23、D = 2，23是素数，那么23转2进制数为10111，颠倒后为11101，11101转换为十进制数后为29，29也是素数，输出Yes。
  2）又比如N = 23,D = 6 , 23转6进制数为35，颠倒后为53，53转10进制数为33，33不是素数，输出No。

代码

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

bool isPrime(const int num)
{
    if(num != 2 && num % 2 == 0 || num == 1)
        return false;
    for(int i = 2;i * i <= num;i++)
    {
        if(num % i == 0)
            return false;
    }
    return true;
}

int main()
{
    int N;
    while(cin >> N)
    {
        if(N < 0)
            break;
        int D;
        cin >> D;
        if(!isPrime(N))
        {
            cout << "No" << endl;
            continue;
        }
        vector<int> digit(100,0);
        int index = 0;
        while( N != 0)
        {
            digit[index++] = N % D;
            N /= D;
        }
        for(int i = 0;i < index;i++)
        {
            N = N * D + digit[i];
        }
        if(isPrime(N))
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
}