1015 Reversible Primes

1015 Reversible Primes

分数 20

全屏浏览

切换布局

作者 CHEN, Yue

单位 浙江大学

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

1.分析

1.要判断原数是否是素数,还要判断反转后的数是否为素数

2.代码

cpp 复制代码
#include<iostream>
#include<cmath>
using namespace std;
const int MAX=1e6+10;
int N,D,re[MAX];
bool check(int x){           //判断是否为素数
    for(int i=2;i<=x/i;i++){
        if(x%i==0) return false;
    }
    return true;
}
int main(){
    while(cin>>N){
        if(N<0) break;        //判断输入结束
        else cin>>D;
        if(N<2||!check(N)) {       //判断原数
            cout<<"No"<<endl;
            continue;
        }
        int num=0;
        while(N){                   //反转
            re[num++]=N%D;
            N/=D;
        }
        for(int i=num-1;i>=0;i--){
            N+=pow(D,num-i-1)*re[i];
        }
        if(N>=2&&check(N)) cout<<"Yes"<<endl;      //判断
        else cout<<"No"<<endl;
    }
    return 0;
}
相关推荐
剪一朵云爱着2 小时前
力扣81. 搜索旋转排序数组 II
算法·leetcode·职场和发展
Molesidy4 小时前
【VSCode】【Clangd】Win下的基于LLVM/Clangd+Clangd插件+MINGW+CMake的VSCode配置C/C++开发环境的详细教程
c++·ide·vscode·clangd·llvm
报错小能手5 小时前
刷题日常 5 二叉树最大深度
算法
码银5 小时前
【数据结构】顺序表
java·开发语言·数据结构
Greedy Alg5 小时前
LeetCode 84. 柱状图中最大的矩形(困难)
算法
Mr_WangAndy5 小时前
C++_chapter13_C++并发与多线程_多线程概念,死锁,unique_lock(),lock_guard()使用
c++·lock·死锁·并发与多线程·unlock·lock_guard·unique_lock
im_AMBER5 小时前
Leetcode 52
笔记·学习·算法·leetcode
小欣加油5 小时前
leetcode 946 验证栈序列
c++·算法·leetcode·职场和发展
神仙别闹5 小时前
基于QT(C++) 实现哈夫曼压缩(多线程)
java·c++·qt