1015 Reversible Primes

1015 Reversible Primes （20 分）

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

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题型分类：进制转换、简单数学问题

题目大意：如果一个正整数是素数，并且他在进制D下的的字符串反转之后，再转换成十进制数也是素数的话，这个数就是reversible prime。

解题思路：分为三步模块化处理：

第一步将给定的数转换成D进制下的字符串（此字符串已经翻转）；

第二步将翻转后的字符串转换成十进制数；

第三步判断原数字和处理后的数是否都为素数。

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#include <cstdio>
#include <string>
#include <algorithm>
#include <cmath>
 
using namespace std;

string convertToDRadix(int N, int D);
int convertToDecimal(string str, int D);
bool isPrime(int N);

int main(int argc, char** argv) {
	int N, D, reverseN;
	string str;
	while(1){
		scanf("%d", &N);
		if(N < 0) break; //负数则表示输入结束 
		scanf("%d", &D);
		str = convertToDRadix(N, D);
		reverseN = convertToDecimal(str, D);
		if(isPrime(N) && isPrime(reverseN)) printf("Yes\n");
		else printf("No\n");
	}	
	return 0;
}

string convertToDRadix(int N, int D){
	string str;
	do{ //注意得到的str已经是反转后的数 
		str += N % D + '0';
		N /= D;
	}while(N);
	return str;
}

int convertToDecimal(string str, int D){
	int sum = 0;
	for(int i = 0; i < str.length(); i++){
		sum = sum * D + str[i] - '0';
	}
	return sum;
}

bool isPrime(int N){
	if(N <= 1) return false;
	int sqr = sqrt(1.0 * N);
	for(int i = 2; i <= sqr; i++){
		if(N % i == 0) return false;
	}
	return true;
}