• PAT-1015 Reversible Primes (20 分) 进制转换+质数


    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");
        }
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Tony100K/p/11758029.html
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