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

    题意

    Language:
    The Luckiest number
    Time Limit: 1000MSMemory Limit: 65536K
    Total Submissions: 7083Accepted: 1886

    Description

    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

    Source

    分析

    [ecause L | frac{8(10^x-1)}{9} \ herefore 9L | 8(10^x-1) \ herefore frac{9L}{gcd(L,8)} | 10^x-1 \ herefore 10^x equiv 1 (mod frac{9L}{gcd(L,8)}) \ ecause a^x equiv 1 (mod n) ightarrow gcd(a,n)=1,x|varphi(n) \ herefore x|varphi(frac{9L}{gcd(L,8)}) ]

    所以枚举即可。时间复杂度(O(sqrt{L} log L))

    代码

    #include<iostream>
    #include<cmath>
    #define rg register
    #define il inline
    #define co const
    template<class T>il T read(){
    	rg T data=0,w=1;
    	rg char ch=getchar();
    	while(!isdigit(ch)){
    		if(ch=='-') w=-1;
    		ch=getchar();
    	}
    	while(isdigit(ch))
    		data=data*10+ch-'0',ch=getchar();
    	return data*w;
    }
    template<class T>il T read(rg T&x){
    	return x=read<T>();
    }
    typedef long long ll;
    
    ll L,d,k,p,s,i,num=0;
    ll gcd(ll x,ll y){
    	return y?gcd(y,x%y):x;
    }
    ll ksc(ll a,ll b,ll c){
    	ll ans=0;
    	while(b){
    		if(b&1) ans=(ans+a)%c;
    		a=a*2%c;
    		b>>=1;
    	}
    	return ans;
    }
    ll ksm(ll a,ll b,ll c){
    	ll ans=1%c;
    	a%=c;
    	while(b){
    		if(b&1) ans=ksc(ans,a,c);
    		a=ksc(a,a,c);
    		b>>=1;
    	}
    	return ans;
    }
    ll phi(ll n){
    	ll ans=n;
    	for(i=2;i*i<=n;++i)
    		if(n%i==0){
    			ans=ans/i*(i-1);
    			while(n%i==0) n/=i;
    		}
    	if(n>1) ans=ans/n*(n-1);
    	return ans;
    }
    ll number(){
    	d=gcd(L,8);
    	k=9*L/d;
    	if(gcd(k,10)!=1) return 0;
    	p=phi(k);
    	s=sqrt((double)p);
    	for(i=1;i<=s;++i)
    		if(p%i==0&&ksm(10,i,k)==1)
    			return i;
    	for(i=s-1;i;--i)
    		if(p%i==0&&ksm(10,p/i,k)==1)
    			return p/i;
    	return 0;
    }
    int main(){
    //	freopen(".in","r",stdin);
    //	freopen(".out","w",stdout);
    	while(read(L)) printf("Case %lld: %lld
    ",++num,number());
    	return 0;
    }
    
  • 相关阅读:
    【动画】看动画轻松理解「Trie树」
    浅析HTTP/2的多路复用
    HTTPS 详解
    PHP写时复制(Copy On Write)
    golang 几种字符串的拼接方式
    正排索引和倒排索引简单介绍
    传值还是传引用
    lvs与nginx区别
    Docker运行操作系统环境(BusyBox&Alpine&Debian/Ubuntu&CentOS/Fedora)
    原创-thanos组件(聚合多个prometheus组件)原理介绍
  • 原文地址:https://www.cnblogs.com/autoint/p/10440135.html
Copyright © 2011-2022 走看看