1015 Reversible Primes (20)(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
解题思路:
素数判断的时候,i<=sqrt(n)这句,我一开始交的时候没有打等于号,错了一个测试点,如果不能理解,那就把73转化成37时写的代码中把10改成d就好了。在反转用do-while比较好,这样等于0的时候就不用单独讨论了。
#include<bits/stdc++.h>
using namespace std;
int Reversible(int n,int d)
{
int sum=0;
do{
sum=sum*d+n%d;
n=n/d;
}while(n!=0);
return sum;
}
int is_prime(int n)
{
if(n==0||n==1)
return 0;
for(int i=2;i<=sqrt(n);i++)
{
if(n%i==0) return 0;
}
return 1;
}
int main(void)
{
int n,d;
while(~scanf("%d%d",&n,&d)&&n>=0)
{
//cout<<Reversible(n,d)<<endl;
if(is_prime(n)&&is_prime(Reversible(n,d)))
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
return 0;
}