poj3696

Posted 2020-11-09 lmjer

4794: The Luckiest Number

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 48  Solved: 8
[Submit][Status][Web Board]

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 positiv
e integers that are a multiple of L and consist of only digit ‘8‘.?
找到一个最小的只含有数字8的十进制正整数，使它为L的倍数，输出其长度

Input

The input consists of multiple test cases. Each test case contains exactly one line containing L(1 
≤ L ≤ 10^12).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

由x个n组成的数可以写成n(10^x-1)/9

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long ll;

ll gcd(ll a,ll b){
   return b?gcd(b,a%b):a;
}

ll getphi(ll n){
   ll ans=n;
   for (ll 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-1)/n;
   return ans;
}

ll quickmod(ll a,ll b,ll p){
    ll ans=1%p;
    for (;b;b>>=1){
        if(b&1) ans=ans*a%p;
        a=a*a%p;
    }
    return ans;
}

int main(){
    ll l;
    ll t=0;
    while(scanf("%lld",&l)&&l!=0){
        ++t;
        ll ans=999999999999;
        ll d=gcd(l,8);
        ll k=9*l/d;
        ll phi=getphi(k);
        if(gcd(k*9/gcd(k,8),10)!=1)
        {
            printf("Case %lld: 0\n",t);
            continue;
        }
        for (ll i=1;i*i<=phi;i++){
            if(phi%i==0){
            if(quickmod(10,i,k)==1%k) ans=min(ans,i);
            else if(quickmod(10,phi/i,k)==1%k) ans=min(ans,phi/i);
            }
        }
        if(ans==999999999999) printf("Case %lld: 0\n",t);
        else printf("Case %lld: %lld\n",t,ans);
    }
return 0;
}