zoukankan      html  css  js  c++  java
  • POJ3696:The Luckiest number(欧拉函数||求某数最小的满足题意的因子)

    Chinese people think of '8' as the lucky digit. Bob also likes digit '8'. Moreover, Bob has his own lucky number L. Now he wants to construct his luckiest number which is the minimum among all positive integers that are a multiple of L and consist of only digit '8'.

    Input

    The input consists of multiple test cases. Each test case contains exactly one line containing L(1 ≤ L ≤ 2,000,000,000).

    The last test case is followed by a line containing a zero.

    Output

    For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the length of Bob's luckiest number. If Bob can't construct his luckiest number, print a zero.

    Sample Input

    8
    11
    16
    0

    Sample Output

    Case 1: 1
    Case 2: 2
    Case 3: 0

    题意:求最小的由8组成的数,是L的倍数。

    思路:由一系列证明得到,ans=phi(N)的满足题意的最小因子。

    关键:    对于X,求其满足题意的最小因子p|X,可以这样求,枚举素因子prime,如果X/prime满足题意,则X=X/prime。

    存疑:我感觉复杂度是根号级别的,但是我看到ppt上说是log级别。 

    #include<cstdio>
    #include<cstdlib>
    #include<iostream>
    using namespace std;
    #define ll long long
    ll gcd(ll a,ll b){ if(b==0) return a;return gcd(b,a%b);}
    ll qmul(ll a,ll x,ll Mod){ll res=0; a%=Mod; while(x){if(x&1) res=(res+a)%Mod;a=(a+a)%Mod;x>>=1;} return res;}
    ll qpow(ll a,ll x,ll Mod){ll res=1; a%=Mod; while(x){if(x&1LL) res=qmul(res,a,Mod); a=qmul(a,a,Mod); x>>=1;} return res;}
    ll phi(ll x)
    {
        ll tx=x,res=x;
        for(int i=2;i*i<=tx;i++){
            if(tx%i==0){
                res-=res/i;
                while(tx%i==0) tx/=i;
            }
        }
        if(tx>1) res-=res/tx;
        return res;
    }
    ll find(ll Mod,ll py) //得到最小因子满足条件 
    {
         ll n=py; ll tp[50][2]; int k=0;
         for(ll i=2;i*i<=n;i++)      //唯一分解  
           if(n%i==0){
             k++; tp[k][0]=i; tp[k][1]=0;
             while(n%i==0){
                n/=i; tp[k][1]++;
             } 
         }
         if(n>1) k++, tp[k][0]=n, tp[k][1]=1; 
         
         for(int i=1;i<=k;i++)
           for(int j=1;j<=tp[i][1];j++)
             if(qpow(10,py/tp[i][0],Mod)==1)
               py/=tp[i][0];
        return py;
    }
    int main()
    {
        ll N,M,Case=0;
        while(~scanf("%lld",&N)&&N){
            printf("Case %lld: ",++Case);
            N=N*9/gcd(N,8);
            if(gcd(N,10)!=1) printf("0
    ");
            else printf("%d
    ",find(N,phi(N)));
        } 
        return 0;
    }
  • 相关阅读:
    log4net的使用
    在asp.net中使用 log4net 笔记
    JQuery插件开发
    使用Visual Studio宏来自动生成代码 [ Visual Studio | 宏 | 自动生成代码 ]
    如何在JBuilder 2006中打Jar包,并生成Exe文件
    如何更改java应用程序标题栏默认图标
    《ASP.NET AJAX程序设计》图书相关资源总索引
    Eclipse汉化的步骤,非常的详细
    家庭和睦、人生平淡也是一种成功
    程序员节诗词
  • 原文地址:https://www.cnblogs.com/hua-dong/p/8504959.html
Copyright © 2011-2022 走看看