1015 Reversible Primes (20分) 甲级真题

可逆素数的D进制判断

最新推荐文章于 2022-09-09 23:13:55 发布

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本文探讨了在给定的D进制数系统中判断一个正整数N是否为可逆素数的问题。可逆素数是指在任何数系统中，其“反转”在该数系统中也是素数的素数。文章提供了判断方法及C++实现代码，通过将数字转换为指定进制并检查其反转数是否为素数来确定。

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

分析：判断输入的数字是否大于0，若大于0则判断是否为素数，若是，则将其转换成d进制，然后从十进制判断是否为素数。是就输出Yes，否则输出No。

#include<iostream>
#include<bits/stdc++.h>
using namespace std;
bool isprime(int n) {
    if (n <= 1) return false;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return true;
}
int main() {
    int num, code, sum;
    for (int i = 0; true; i++) {
        sum = 0;
        cin >> num;
        if (num > 0) {
            cin >> code;
            if (!isprime(num)) { cout << "No\n"; continue; }
            while (num) {
                sum = sum * code + num % code;
                num /= code;
            }
            if (isprime(sum)) cout << "Yes\n";
            else cout << "No\n";
        }
        else break;
    }
    return 0;
}