zoukankan      html  css  js  c++  java
  • UVA 10006

      Carmichael Numbers 

    An important topic nowadays in computer science is cryptography. Some people even think that cryptography is the only important field in computer science, and that life would not matter at all without cryptography.  Alvaro is one of such persons, and is designing a set of cryptographic procedures for cooking paella. Some of the cryptographic algorithms he is implementing make use of big prime numbers. However, checking if a big number is prime is not so easy. An exhaustive approach can require the division of the number by all the prime numbers smaller or equal than its square root. For big numbers, the amount of time and storage needed for such operations would certainly ruin the paella.

    However, some probabilistic tests exist that offer high confidence at low cost. One of them is the Fermat test.

    Let a be a random number between 2 and n - 1 (being n the number whose primality we are testing). Then, n is probably prime if the following equation holds: 

     

    egin{displaymath}a^n mod n = aend{displaymath}

     

    If a number passes the Fermat test several times then it is prime with a high probability.

    Unfortunately, there are bad news. Some numbers that are not prime still pass the Fermat test with every number smaller than themselves. These numbers are called Carmichael numbers.

    In this problem you are asked to write a program to test if a given number is a Carmichael number. Hopefully, the teams that fulfill the task will one day be able to taste a delicious portion of encrypted paella. As a side note, we need to mention that, according to Alvaro, the main advantage of encrypted paella over conventional paella is that nobody but you knows what you are eating.

     

    Input 

    The input will consist of a series of lines, each containing a small positive number  n  (  2 <  n  < 65000). A number  n  = 0 will mark the end of the input, and must not be processed.

     

    Output 

    For each number in the input, you have to print if it is a Carmichael number or not, as shown in the sample output.

     

    Sample Input 

    1729
    17
    561
    1109
    431
    0
    

     

    Sample Output 

    The number 1729 is a Carmichael number.
    17 is normal.
    The number 561 is a Carmichael number.
    1109 is normal.
    431 is normal.
    

     

    题意:判断一个数是不是Carmichael数。 

    如果一个数不是素数,且对于任意的2< a <n满足方程 egin{displaymath}a^n mod n = aend{displaymath},则称n是Carmichael数;否则n就不是Carmichael数。

    这个题的关键是求快速幂。

     

    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    #define LL long long
    int a[66000];
    void judge_prime() /*筛法求素数*/
    {
        int i,j,m=sqrt(65010+0.5);
        memset(a,0,sizeof(a));
        for(i=2;i<=m;i++)
        {
            if(!a[i])  /*素数为0*/
            {
                for(j=i*i;j<65010;j+=i)
                    a[j]=1; /*非素数为1*/
            }
        }
    }
    LL pow_mod(LL a,LL n,LL m) /*递归求快速幂*/
    {
        if(n==0) return 1;
        LL x=pow_mod(a,n/2,m);
        LL ans=x*x%m;
        if(n%2==1) ans=ans*a%m;
        return ans;
    }
    int main()
    {
        judge_prime();
        LL i,n;
        bool flag;
        while(~scanf("%lld",&n)&&n)
        {
            if(!a[n])
            {
                printf("%lld is normal.
    ",n);
                continue;
            }
            flag=true;
            for(i=2;i<n;i++)
            {
                if(pow_mod(i,n,n)!=i)
                {
                    flag=false;
                    break;
                }
            }
            if(flag)
                printf("The number %lld is a Carmichael number.
    ",n);
            else
                printf("%lld is normal.
    ",n);
        }
        return 0;
    }



  • 相关阅读:
    深入理解虚拟机、容器和Hyper技术
    Hyper:基于Hypervisor的容器化解决方案
    iSCSI 在Linux下的模拟实验
    C-RAN
    spring mvc
    DUBBO
    ajax跨域请求 小栗子 jsonP
    freemarker 页面静态化 简单小栗子
    ajax上传图片监听
    代码优化 粘贴
  • 原文地址:https://www.cnblogs.com/dyllove98/p/3235481.html
Copyright © 2011-2022 走看看