1015. Reversible Primes (20)

1015. Reversible Primes (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题型：素数

题目大意：如果一个数本身是素数，而且在d进制下反转后的数在十进制下也是素数，就输出Yes，否则就输出No
分析：判断输入是否为负数，判断n是否为素数，把n转换为d进制再反过来转换为10进制，判断是否为素数

#include<iostream>
#include<cmath>
using namespace std;
int n,radix;
bool prime(int num)
{
if(num<2) return false;//之前这一句没写，导致一个案例没有通过，还是的细心啊
if(num==2 || num==3)
return true;
if(num%2==0)//去掉偶数
return false;
for(int i=3;i<=sqrt(num);i+=2)//遍历到到这个数的二次根即可
{
if(num%i==0)
return false;
}
return true;
}
int change(int num)
{
int ch[50];
int index=0;
while(num!=0)
{
ch[index++]=num%radix;
num/=radix;
}
num=0;
for(int i=0;i<index;++i)
{
num=num*radix+ch[i];
}
return num;
/*还可以这样子写
int n=0;
while(num!=0)
{
n=radix*n+num%radix;
num/=rasix;
}
return n;
*/
}
int main()
{
//freopen("in.txt","r",stdin);
cin>>n;
while(n>0)
{
cin>>radix;
if(prime(n)==false)
{
cout<<"No"<<endl;
}
else
{
if(prime(change(n))==true)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
cin>>n;
}
return 0;
}