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 -2 Sample Output: Yes Yes No 题意： 判断N是不是素数，然后将N转换成D进制的数，然后翻转，然后转成10进制，然后判断结果和N是不是都是素数。 如23 2 转换成2进制是10111 反转是11101 转换成十进制是29 因为29和23都是素数，所以输出yes ps:英语不行，做个题目都费劲 #include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> #include<vector> using namespace std; int prime[100004]; void isprime() { memset(prime,1,sizeof(prime)); prime[1]=prime[0]=0; for(int i=2;i<=100004;i++) { if(prime[i]) { for(int j=i*2;j<=100004;j+=i) { prime[j]=0; } } } } int main() { isprime(); int n;int r; while(cin>>n) { if(n<0)break; cin>>r; int a=n; vector<int>v; while(n/r) { v.push_back(n%r); n/=r; } v.push_back(n); int b=0; for(int i=0;i<v.size();i++) { b=b*r+v[i]; } if(prime[a]&&prime[b]) { cout<<"Yes"<<endl; } else{ cout<<"No"<<endl; } } return 0; }