The Luckiest number（hdu2462）

The Luckiest number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1163 Accepted Submission(s): 363


[align=left]Problem Description[/align]
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'.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

8

11
16
0

[align=left]Sample Output[/align]

Case 1: 1
Case 2: 2
Case 3: 0
思路：欧拉函数；
其实这题和hdu3307，基本一样，只不过这个推下。
设f
,表示n位全是8的数，那么f
=10*f[n-1]+8,那么构造等比数列f
+(8/9)=10*(f[n-1]+(8/9));
那么f
= (8+8/9)*(10)^(n-1)-8/9;f
= (8/9)*（(10)^n-1）;那么就是要求最小的n使f
%L=0;
那么(8/9)*（10^n-1）=k*L;
8/gcd(8,L)*（10^n-1）=9*k*L/(gcd(8,L));
化简为8/gcd(8,L)*（10^n）%（9*L/(gcd(8,L))）=8/gcd（8，L）;
8/gcd（8，L）与（9*L/(gcd(8,L))互质可以消去，的10^n%（9*L/(gcd(8,L))）=1;
那么用另模数为m，10^n%（m）=1;

m和10必定互质，否则无解。

于是根据欧拉定理，10^(Euler(m)) = 1(mod m) 。由于题目要求最小的解，解必然是Euler(m)的因子。

需要注意的是，对于10^x，由于m太大，直接快速幂相乘的时候会超long long

这题我开始用baby-step,超时了；

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<queue>
#include<set>
#include<math.h>
#include<map>
using namespace std;
typedef long long LL;
pair<LL,LL>exgcd(LL n,LL m);
LL gcd(LL n,LL m);
LL quick(LL n,LL m,LL mod);
LL mul(LL n, LL m,LL p);
int  slove(LL n);
LL phi(LL n);
bool prime[1000005];
int ans[1000005];
LL fen[1000005];
int main(void)
{
LL n;
int i,j;
int cn = 0;
for(i = 2; i <= 1000; i++)
{
if(!prime[i])
{
for(j = i; (i*j) <= 1000000; j++)
{
prime[i*j] = true;
}
}
}
for(i = 2; i <= 1000000; i++)
{
if(!prime[i])
{
ans[cn++] = i;
}
}
//printf("%d\n",cn);
int __ca = 0;
while(scanf("%lld",&n),n!=0)
{
LL gc = gcd(8,n);
n = 9*n/gc;
LL oula = phi(n);
LL x = gcd(n,10);//printf("%lld\n",n);
//printf("%lld\n",x);
printf("Case %d: ",++__ca);
if(x!=1)
{
printf("0\n");
}
else
{
int k = slove(oula);
//printf("%d\n",k);
for(i = 0;i < k;i++)
{
LL akk =quick(10,fen[i],n);
if(akk==1)
{
break;
}
}//printf("%d\n",10);
printf("%lld\n",fen[i]);
}
}
return 0;
}
int  slove(LL n)
{   int cn = 0;int i,j;
for(i = 1;i < sqrt(1.0*n);i++)
{
if(n%i==0)
{
if(n/i==i)
{
fen[cn++] = i;
}
else
{
fen[cn++] = i;
fen[cn++] = n/i;
}
}
}
sort(fen,fen+cn);
return cn;
}
LL phi(LL n)
{
int f = 0;
bool flag = false;
LL ask =n;
while(n>1)
{
while(n%ans[f]==0)
{
if(!flag)
{
flag = true;
ask/=ans[f];
ask*=ans[f]-1;
}
n/=ans[f];
}
f++;
flag = false;
if((LL)ans[f]*(LL)ans[f]>n)
{
break;
}
}
if(n > 1)
{
ask/=n;
ask*=(n-1);
}
return ask;
}
pair<LL,LL>exgcd(LL n,LL m)
{
if(m==0)
return make_pair(1,0);
else
{
pair<LL,LL>ak = exgcd(m,n%m);
return make_pair(ak.second,ak.first-(n/m)*ak.second);
}
}
LL gcd(LL n,LL m)
{
if(m==0)
return n;
else return gcd(m,n%m);
}
LL quick(LL n,LL m,LL mod)
{
LL ak = 1;
n %= mod;
while(m)
{
if(m&1)
ak =mul(ak,n,mod);
n = mul(n,n,mod);
m>>=1;
}
return ak;
}
LL mul(LL n, LL m,LL p)
{
n%=p;
m%=p;
LL ret=0;
while(m)
{
if(m&1)
{
ret=ret+n;
ret%=p;
}
m>>=1;
n<<=1;
n%=p;
}
return ret;
}