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
题意:给定一个正整数n,和一个进制d,如果n在10进制下是素数,并且将其d进制逆转之后再转换成10进制数,仍然是素数,则输出Yes,否则输出No。
思路:自己编写两个进制转换函数,以及素数判断函数按要求操作即可。
参考代码:
#include<cstdio>
#include<iostream>
#include<string>
#include<cmath>
using namespace std;
bool judge(int n)
{
if(n<=1) return false;
int sqrtN=(int)sqrt(1.0*n);
for(int i=2;i<=sqrtN;i++)
{
if(n%i==0)
return false;
}
return true;
}
string intToA(int n,int radix)
{
string ans="";
do{
ans+=n%radix+'0';
n/=radix;
}while(n!=0); //使用do{}while()以防止输入为0的情况
return ans; //此时的字符串正好存储逆置后的结果
}
int main()
{
int n,radix;
while(cin>>n,n>=0)
{
cin>>radix;
string s;
if(!judge(n)) {
printf("No\n");
continue;
}
s=intToA(n,radix);
n=0;
for(int i=0;i<s.size();i++)
{
n=n*radix+s[i]-'0';
}
if(judge(n)) printf("Yes\n");
else printf("No\n");
}
return 0;
}