【PAT】【Advanced Level】1015. Reversible Primes (20)

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本文介绍了一种算法，用于判断一个正整数N在特定基数D下是否为可逆素数。通过分解、翻转并模拟该过程，利用C++实现了一个高效的解决方案。

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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:

Sample Output:

Yes
Yes
No

原题链接：

https://www.patest.cn/contests/pat-a-practise/1015

思路：

按要求分解、翻转、模拟

坑点：

不要忘记判断原数是否为0

CODE：

#include<iostream>
using namespace std;
bool pri(int n)
{
    if (n==0 || n==1) return false;
    for (int i=2;i*i<=n;i++)
    {
        if (n%i==0) return false;
    }
    return true;
}
int rev(int n,int ra)
{
    string re="";
    while (n!=0)
    {
        re+=('0'+n%ra);
        n/=ra;
    }
    int rev=0;
    for (int i=0;i<re.length();i++)
    {
        rev*=ra;
        rev+=(re[i]-'0');
    }
    return rev;
}

int main()
{
    int n;
    int ra;
    cin>>n;
    while (n>=0)
    {
        cin>>ra;
        if (pri(n)==0)
        {
            cout<<"No"<<endl;
        }
        else
        {
            if (pri(rev(n,ra)))
            {
                cout<<"Yes"<<endl;
            }
            else
            {
                cout<<"No"<<endl;
            }
        }
        cin>>n;
    }
    return 0;
}