1015 Reversible Primes (20 分)
标签：not gre specific namespace strong 根据 str 判断 mat
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 (<) and D (1), you are supposed to tell if N is a reversible prime with radix D.
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.
For each test case, print in one line Yes if N is a reversible prime with radix D, or No if not.
73 10
23 2
23 10
-2
Yes
Yes
No
题目大意：
一开始理解错了题意。。。。
数字N在D进制下是不是双重素数。双重素数是本身和倒数皆为素数的数。
实现：判断N是否为素数。如果不是，输出No，否则将该数在D进制下倒过来再化为十进制数，判断是否为素数。如果是，输出Yes，否则输出No.
复习判断素数知识点 注意 ‘ = ’ ！！！
#include<bits/stdc++.h> using namespace std; bool prime(int x){ if(x==0||x==1){ return false; } if(x==2){ return true; } for(int i=2;i<=sqrt(x);i++){//这个地方忘记了=号！！！ if(x%i==0){ return false; } } return true; } int main() { int a; int d; while(cin>>a) { if(a<0){ break; } cin>>d; //先判断本身是不是素数 if(!prime(a)){ cout<<"No"<<endl; continue; } //根据相应地进制转 string s=""; int x; while(a){ s+=char(a%d+‘0‘); a=a/d; } //cout<<s<<endl; //反向再把它从d进制转成10进制 int l = s.length(); x=0; for(int i=0;i<l;i++){ x=x*d+s[i]-‘0‘; } //cout<<x<<endl; if(prime(x)){ cout<<"Yes"<<endl; } else{ cout<<"No"<<endl; } } return 0; }
PAT 甲级 1015 Reversible Primes (20 分) （进制转换和素数判断(错因为忘了=)）
标签：not gre specific namespace strong 根据 str 判断 mat
原文地址：https://www.cnblogs.com/caiyishuai/p/11291829.html
