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  • A1015. Reversible Primes (20)

     

    时间限制
    400 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    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

    #include <stdio.h>
    #include <stdlib.h>
    #include <iostream>
    #include <string.h>
    #include <string>
    #include <math.h>
    #include <algorithm>
    using namespace std;
    
    bool isPrime(int n)
    {
    	if(n<=1)return false;
    	int sqr=sqrt(1.0*n);
    	for(int i=2;i<=sqr;i++)
    	{
    		if(n%i==0)return false;
    	}
    	return true;
    } 
    
    
    int main(){
        int n,radix;
        int a[100]={0};
        //scanf("%d %d",&n,&radix);
        while(scanf("%d",&n)!=EOF)
        {
        	if(n<0)break;
        	scanf("%d",&radix);
        	if(isPrime(n)==false)
        	{
        		printf("No
    ");
        	}
        	else
        	{
        		int len=0;//进制转换
    			do{
    				a[len++]=n%radix;
    				n/=radix;
    			}while(n!=0);
    			for(int i=0;i<len;i++)
    			{
    				n=n*radix+a[i];
    			} 
    			if(isPrime(n)==false)printf("No
    ");
    			else printf("Yes
    ");
        	}
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/ligen/p/4304859.html
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