1015 Reversible Primes

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最新推荐文章于 2024-06-08 10:35:18 发布

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本文介绍了一种判断可逆素数的方法，通过将一个给定的十进制数转换为指定进制，再进行反转并转回十进制，最后验证原始数及其反转数是否均为素数。

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题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805495863296000

题目描述：

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

解题思路：

1、输入 n与d，n代表一个十进制数，d代表进制。

2、将n化为d进制数nd，然后将nd反转dn，再将dn转换为十进制数:ren

3、判断n和ren是否均为素数，是输出“Yes”，不是输出"No"

4、举例：

输入 23 2：23的2进制表示：10111，转置以后 11101 ,化为10进制数29，23与29均为素数，输出‘Yes’

解题关键：获取素数；

解题代码

#include <iostream>
#include<bits/stdc++.h>
//https://pintia.cn/problem-sets/994805342720868352/problems/994805495863296000
using namespace std;
const int MAXSIZE = 100001;
int prime[MAXSIZE];
//将num转换为d进制，然后倒置再变为10进制
int getData(int num,int d){
    int t = 1;
    int n = 0;
    int sumNum = 0;
    int di[100];
    while(num>0){
        di[sumNum++] = num%d;
        num/=d;
    }
    for(int i = sumNum-1 ;i>=0;i--){
        n += (t*di[i]);
        t*=d;
    }
    return n;
}
//获取素数
void initPrime(){
    int i,j;
    for(i =2;i<MAXSIZE;i++){
        prime[i] = 1;
    }
    for(i = 2;i<MAXSIZE;i++){
        if(prime[i]==1){
            for(j = i*2;j<MAXSIZE;j+=i){
                prime[j] = 0;
            }
        }
    }
}
int main()
{
    int n,d;
    initPrime();
    while(1){
        cin>>n>>d;
        if(n<0){
            break;
        }
        if(prime[n]&&prime[getData(n,d)]){
            cout<<"Yes"<<endl;
        }else{
            cout<<"No"<<endl;
        }
    }
}