PAT (Advanced Level) Practice 1015 Reversible Primes (20分) （进制转换+素数判断）

1.题目

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​5​​) 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

2.题目分析

先看给出的第一个数是不是素数，不是就输出NO，是的话再将这个数按照后面的进制要求转换为相应进制下的数，将这个数reverse一下，再转为十进制看是不是素数，是的话就是yes

3.代码

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
bool isprime(int t)
{
	if (t == 0 || t == 1)return false;
	for (int i = 2; i <= sqrt(t); i++)
		if (t%i == 0)return false;
	return true;
}
int changea(string a,int d)//其它进制转十进制
{
	int num = 0;
	for (int i = 0; i < a.length(); i++)
		num += (a[i] - '0')*pow(d, a.length() - i - 1);
	return num;
}
string changeb(int num,int d )//十进制转其它进制
{
	string answer = "";
	int out[110];
	int s = 0;
	while (num > 0)
	{
		out[s++] = num%d;
		num /= d;
	}
	for (int i = s - 1; i >= 0; i--)
		answer += out[i] + '0';
	reverse(answer.begin(), answer.end());
	return answer;
}
int main()
{
	string a, b;
	while (1)
	{
		cin >> a;
		if (stoi(a) < 0)return 0;
		cin >> b;
		if (!isprime(stoi(a))) { printf("No\n"); continue; }
		if(!isprime(changea(changeb(stoi(a),stoi(b)),stoi(b)))) { printf("No\n"); continue; }
		printf("Yes\n");
	}
}