title
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
solution
题目大意：判断是否反素数，即：本身是素数，在D进制下的reverser数还是素数
注意以下几点：
- 判断素数，记得<=1的都是非素数返回false
- 看题啊！！！
#include<iostream>
#include<string>
#include<stdio.h>
using namespace std;
bool isPrime2(long long);
bool isPrime(long long a, int radix) {
string s = "";
while (a > 0) {
s += (a%radix+'0');
a /= radix;
}
long long b = 0;
for (int i = 0; i < s.length(); i++) {
b = b * radix + (s[i] - '0');
}
//cout << "--------" << a<<" --- "<<b << endl;
return isPrime2(b);
}
bool isPrime2(long long n) {
if (n <= 1)
return false;
for (int i = 2; i*i <= n; i++) {
if (n%i == 0) {
return false;
}
}
return true;
}
int main() {
long long a;
int radix;
while (scanf(" %lld", &a)==1&&a>0) {
scanf(" %d", &radix);
if (isPrime2(a)&&isPrime(a, radix)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
}
}