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
注意读题,首先他得是一个质数才行,这个先决条件我一开始没有判断,另外判断质数时要注意1,2,3和负数的情况
#include<bits/stdc++.h>
#define de(x) cout<<#x<<" "<<(x)<<endl
#define each(a,b,c) for(int a=b;a<=c;a++)
using namespace std;
const int maxn=50+5;
int buf[maxn];
bool isprime(int x)
{
if(x<=1)return false;
bool flag=true;
if(x==2)return true;
if(x==3)return true;
for(int i=2;i*i<=x;i++)
{
if(x%i==0)
{
flag=false;
break;
}
}
return flag;
}
int main()
{
//de(isprime(73));
int n,r;
while(true)
{
cin>>n;
if(n<0)break;
cin>>r;
if(n==1||n==0)
{
puts("No");
continue;
}
if(!isprime(n))///读题的问题,首先得判断他是不是一个质数才行,
{
puts("No");
continue;
}
int len=0;
while(n>0)
{
buf[len++]=n%r;
n/=r;
}
//de(len);
int sum=0;
len--;
for(int i=0;i<=len;i++)
{
sum+=buf[i]*pow(r,len-i);
}
//de(sum);
if(isprime(sum))puts("Yes");
else puts("No");
}
}
原文地址:https://www.cnblogs.com/Tony100K/p/11758029.html
时间: 2024-10-17 21:10:03