#POJ3696# The Luckiest number

2019-11-11 01:53:41

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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'.InputThe 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.OutputFor 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

811160Sample OutputCase 1: 1Case 2: 2Case 3: 0
题意：
给出一个数L(L<2,000,000,000)，找一个最小的L的倍数，使其乘积的每个数,字都是8，求这个乘积的位数。
设此乘积位数为x，倍数为k
则有
(10 ^ x - 1) * 8 / 9 = k * L
      10 ^ x - 1 = 9 / 8 * k * L
令 m = 9 * L / gcd (L , 8), y = k * gcd (L , 8) / 8
若此L有解，则
1：满足上式
2：k为整数
3：y为整数
假设1，2成立，则3若成立，L即有解
则有
(10 ^ x - 1) % m = 0
 10 ^ x % m = 1
由欧拉公式得： 若上式有解，则必有10与m互素，且x = φ(m)
φ(m)肯定为L有解的一个倍数，但不一定是最小的，x的最小值肯定为φ(m)的约数
于是穷举约数检查是否符合10 ^ x % m = 1，找到最小x
注意：
1：枚举时只枚举到sqrt(φ(m))再倒回去通过前半段算出后半段即可，节省时间
2：此题检验时由于是10的指数，容易溢出，所以快速幂相乘时改成了加法做乘法，倍增加法，原理同倍增快速幂
Code:
Status Accepted
Time 156ms
Memory 1728kB
Length 1063
Lang C++
Submitted 2017-02-07 17:20:08
Shared 
#include<iostream>#include<cstdio>#include<cmath>using namespace std;typedef long long LL;LL gcd(LL A,LL B){return B==0?A:gcd(B,A%B);}LL mul(LL a, LL b, LL mod){//考虑换成m个10相加，倍增+mod	LL rt=0;	while(b){		if(b&1)	rt=(rt+a)%mod;		a=(a<<1)%mod;		b>>=1;	}	return rt;}LL qmul(LL a, LL b, LL mod){	LL s=1,tmp=a;	while(b){		if(b&1)	s=mul(s,tmp,mod);		tmp=mul(tmp,tmp,mod);		b>>=1;	}	return s;}LL Euler(LL n){	LL m=(int)sqrt(n+0.5);	LL rt=n;	for(LL i=2; i<=m; ++i)if(n%i==0){		rt=rt/i*(i-1);		while(n%i==0)n/=i;	}	if(n>1)rt=rt/n*(n-1);	return rt;}int main(){	int tot=0,d;	LL L,m;	while(scanf("%I64d",&L)&&L){		m=9*L/gcd(L,8);		printf("Case %d: ",++tot);		d=gcd(10,m);		if(d==1){			LL phi=Euler(m);			LL ans=phi;			LL p=sqrt((double)phi);			bool kk=0;			for(int i=1; i<=p; ++i)if(phi%i==0&&qmul(10,i,m)==1){				ans=i,kk=1;				break;			}			if(!kk){				for(int i=p; i>=2; --i)if(phi%i==0&&qmul(10,phi/i,m)==1){					ans=phi/i,kk=1;					break;				}			}			printf("%I64d/n",ans);		}		else {printf("0/n");}	}}