(PAT 1015) Reversible Primes (进制转换+判断素数)

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

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

解题思路:

将给出的10进制数给出后转为对应的进制数,然后判断这个数以及这个数的对称是不是素数

进制转换法为取余法:

#include 
#include 
#include 
#include 
using namespace std;

bool isPrime(int num) {
	if (num <= 1) return false;
	for (int i = 2; i*i <= num; ++i) {
		if (num%i == 0) return false;
	}
	return true;
}
int BTOD(string num, int Rad) {
	int res = 0;
	for (int i = 0; i < num.length(); ++i) {
		res += (num[i] - '0') * pow(Rad, i);
	}
	return res;
}
void DToB(int num,int Rad) {
	int index = 0;
	string input;
	int temp = num;
	while (temp) {
		input += to_string(temp % Rad);
		temp = temp / Rad;
	}
	
	string newstring;
	for (int i = input.length()-1; i >= 0; --i) {
		newstring += input[i];
	}
	if (isPrime(BTOD(input,Rad)) && isPrime(BTOD(newstring,Rad))) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}
}
int main() {
	
	int TNum , Dnum;
	scanf("%d", &TNum);
	while (TNum > 0) {
		scanf("%d", &Dnum);
		DToB(TNum, Dnum);
		scanf("%d", &TNum);
	}
	system("PAUSE");
	return 0;
}