PAT甲级1015 素数

题目

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

---

题解

更新：AC版本

#include<iostream>
#include<cmath>
using namespace std;
bool isPrime(int n){
  int sqr=int(sqrt(n));
  if(n<=1) return false;
  for(int i=2;i<=sqr;i++){
    if(n%i==0) return false;
  }
  return true;
}
int reverse(int n,int d){
  int arr[100];
  int len=0;
  int sum=0;
  int r=1;
  while(n!=0){
      arr[len++]=n%d;
      n/=d;
    }
  for(int i=len-1;i>=0;i--){
     
      sum+=arr[i]*r;
      r*=d;
    }
  
  return sum;
  
}
int main(){
  int n,d,ans;
  while(scanf("%d",&n)!=EOF){
    if(n<=0)  break;//题目要求为负数，终止整个循环
    scanf("%d",&d);
    if(isPrime(n)==false){
      printf("No\n");
      continue;//提前结束此次循环
    }
   ans=reverse(n,d);
   if(isPrime(ans)) printf("Yes\n");
   else printf("No\n");
  }
  return 0;
}

---

未AC版本

#include<iostream>
#include<cmath>
using namespace std;
bool isPrime(int n){
  int sqr=int(sqrt(n));
  if(n<=1) return false;
  for(int i=2;i<=sqr;i++){
    if(n%i==0) return false;
    
  }
  return true;
}
int main(){
  int n,d;
  int sum=0;
  int r=1;
  
  while(scanf("%d",&n)!=EOF){
    if(n<=0)  break;//题目要求为负数，终止整个循环
    scanf("%d",&d);
    if(isPrime(n)==false){
      printf("No\n");
      continue;//提前结束此次循环
    }
   
  int arr[100];
  int len=0;
    while(n!=0){
      arr[len++]=n%d;
      n/=d;
    }
    
    for(int i=len-1;i>=0;i--){
      sum+=arr[i]*r;
      r*=d;
    }//主要就是这边的原因，使得两测试点过不去
    
   if(isPrime(sum)) printf("Yes\n");
   else printf("No\n");
  }
  return 0;
}

一开始我在想，D是否就是N的进制，直接把N倒过来，再转10进制。然后发现第二行23后跟着2，这时必须果断猜测，是要把10进制的N转成D进制再倒过来，再转10进制。

我觉得可能要更注意变量要处在合适的位置，以及函数功能方面，能分模块写就分模块写。