zoukankan      html  css  js  c++  java
  • POJ3696 The Luckiest Number

    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

    题解:这种题目一般都是 转化为: num*(10^x-1)/9= L*k

    8*(10^x-1)=9L*k

    设 d=gcd(9L,8)=gcd(8,L)

    p=8/d,q=9L/d;

    则: p*(10^x-1) q*k;

    因为q,p互质,则q|(10^x-1) ,p|k

    则     10^x-1=0(mod q)

    10^x =1(mod q)

    10^x=1(mod 9L/d)

    当q与10互质时10^(oula(q))=1(mod  q)

    因此,字需要枚举其因子即可;
    参考代码:

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<algorithm>
     4 #include<cstring>
     5 using namespace std;
     6 #define clr(a,val) memset(a,val,sizeof(a))
     7 typedef long long ll;
     8 const int INF=0x3f3f3f3f; 
     9 ll L,fac[1010]={0};
    10 inline ll phi(ll x)
    11 {
    12     ll p=x,s=x;
    13     for(ll i=2;i*i<=s;++i)
    14         if(!(x%i))
    15         {
    16             p=p/i*(i-1);
    17             while(!(x%i)) x/=i;
    18         }
    19     if(x>1) p=p/x*(x-1);
    20     return p;
    21 }
    22 
    23 inline void find_factor(ll x)
    24 {
    25     ll s=x;
    26     fac[0]=0;
    27     for(ll i=2;i*i<=s;++i)
    28         if(!(x%i))
    29         {
    30             fac[++fac[0]]=i;
    31             while(!(x%i)) x/=i;
    32         } 
    33     if(x>1) fac[++fac[0]]=x;
    34 }
    35 
    36 inline ll mult(ll a,ll b,ll mod)
    37 {
    38     a%=mod; b%=mod;
    39     ll s=a,sum=0;
    40     while(b)
    41     {
    42         if(b&1)
    43         {
    44             sum+=s;
    45             if(sum>=mod) sum-=mod;
    46         }
    47         b>>=1;s<<=1;
    48         if(s>=mod) s-=mod;
    49     }
    50     return sum;
    51 }
    52 ll power(ll a,ll b,ll mod)
    53 {
    54     ll s=a,sum=1;
    55     while(b)
    56     {
    57         if(b&1) sum=mult(sum,s,mod);
    58         b>>=1;s=mult(s,s,mod);
    59     }
    60     return sum;
    61 }
    62 ll gcd(ll a,ll b) {return b==0? a:gcd(b,a%b);}
    63 int main()
    64 {
    65     int t=0;
    66     while(~scanf("%lld",&L) && L)
    67     {
    68         ++t;
    69         ll m=L/gcd(L,8)*9,p=phi(m),x=p;
    70         if(gcd(m,10)!=1) {printf("Case %d: 0
    ",t);continue;}
    71         find_factor(p);
    72         for(int  i=1;i<=fac[0];++i)
    73         {
    74             while(1)
    75             {
    76                 x/=fac[i];
    77                 if(power(10,x,m)!=1)
    78                 {
    79                     x*=fac[i];
    80                     break;
    81                 }
    82                 else if(x%fac[i]) break;
    83             }
    84         }
    85         printf("Case %d: %lld
    ",t,x);
    86     }
    87     return 0;
    88 } 
    View Code
  • 相关阅读:
    网盘无法单独同步某个文件的解决方法
    编译cubieboard android 源码过程详解之(七):lichee build
    cb-A10系统优化之(一):去除自启动软件
    ubuntu 使用
    JS——数组中push对象,覆盖问题,每次都创建一个新的对象
    Node.js中npm常用命令大全
    Vue style里面使用@import引入外部css, 作用域是全局的解决方案
    5大浏览器内核和主要代表
    IE调试网页之三:使用 F12 工具控制台查看错误和状态 (Windows)
    div拖拽到iframe上方 导致 缩放和拖拽的不平滑和鼠标事件未放开 解决方法
  • 原文地址:https://www.cnblogs.com/csushl/p/10388823.html
Copyright © 2011-2022 走看看