原题传送门
题意：给定数10进制数N和进制D，将N转化成D进制，再将D进制的数反转，最后转化为10进制数M，判断N和M是否都是质数。 #include <iostream> using namespace std; int isPrime(int n) { if(n < 2) return 0; for(int i = 2; i*i <= n; ++i) { if(n%i == 0) return 0; } return 1; } int change(int n, int d) { int b[100], count = 0; while(n) { b[count] = n%d; count++; n /= d; } // 得到的b[]正序就是转换后的数的逆序 int decimal = 0; for(int i = count-1; i >= 0; --i) { int temp = 1; for(int j = 0; j < count-1-i; ++j) { temp = temp * d; } decimal = decimal + b[i] * temp; } return decimal; } int main(int argc, const char * argv[]) { int N, D; while(cin >> N) { if(N < 0) break; cin >> D; int re = change(N, D); if(isPrime(N) && isPrime(re)) cout << "Yes"<<endl; else cout << "No"<<endl; } return 0; }附原题：
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 (< 105) 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 -2Sample Output: Yes Yes No