1015 Reversible Primes （20 分）（进制转换，素数）

最新推荐文章于 2023-02-09 12:14:14 发布

小火汁猛猛

最新推荐文章于 2023-02-09 12:14:14 发布

阅读量161

点赞数 1

CC 4.0 BY-SA版权

分类专栏： c++ 甲级文章标签： PAT

本文链接：https://blog.youkuaiyun.com/qq_42068614/article/details/86545859

c++ 同时被 2 个专栏收录

115 篇文章

订阅专栏

甲级

92 篇文章

订阅专栏

本文介绍了一种算法，用于检测给定的正整数N是否为特定基数D下的可逆素数。可逆素数是指在一个数制下，其数字反转后的数仍然是素数的数。文章详细解释了算法的工作原理，包括将数字转换为指定基数下的表示，然后检查该数及其反转数是否均为素数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

) 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

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <string.h>
#include <cstdio>
using namespace std;
bool isprime(int num){
    if(num==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,d;
    while(cin>>n){
        if(n<0) break;
        cin>>d;
        if(!isprime(n)){
            cout<<"No\n";
            continue;
        }
        int len=0,arr[100];
        while(n!=0){            //转换为d进制数
            arr[len++]=n%d;
            n/=d;
        }
        for(int i=0;i<len;i++)  //倒置相加转换为十进制
            n=arr[i]+n*d;
        if(isprime(n)) cout<<"Yes\n";
        else cout<<"No\n";
    }
    return 0;
}