技術標籤：PAT甲級# 數論PATPAT甲級

立志用最少的程式碼做最高效的表達

PAT甲級最優題解——>傳送門

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

題意：如果數n是素數， 將n轉化成d進位制數，翻轉後在轉換回來， 判斷是否為負數。

舉例：輸入23 2

1. 23為素數，繼續
2. 23轉為二進位制：10111
3. 10111翻轉：11101
4. 11101轉回十進位制：29
5. 29為素數，輸出Yes

本題用了素數的快速判定。

具體原理見部落格——>傳送門

#include <bits/stdc++.h>
using namespace std;

bool isPrime(int n) {
	if(n == 0 || n == 1) return false;
	if
 
(n == 2) return true;
	
	if(n%6 != 1 && n%6 != 5) return false;	//不在6的周圍一定不是 
	int temp = sqrt(n);						//在6的周圍也可能不是 
	for(int i = 5; i <= temp; i += 6) 		
		if(n%i==0 || n%(i+2)==0) return false;
	return true;
}

int main() {
	int n, d; while(cin>>n && n>=0 && cin>>d) {
		int n1 = n;
		string s;
		while(n1) {
			s += (char)(n1%d+'0');
			n1 /= d;
		}
		int sum = 0, siz = s.size()-1;
		for(auto i : s) 
			sum += (i-'0')*pow(d, siz--);
		
		cout << ((isPrime(n) && isPrime(sum)) ? "Yes" : "No") << '\n';
	} 
    return 0;
}

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