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Solution

懶得寫啦

Code

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #define Rg register
 5 #define go(i,a,b) for(Rg int i=a;i<=b;i++)
 6 #define yes(i,a,b) for(Rg int i=a;i>=b;i--)
 7 #define ll long long
 8 using namespace std;
 9 int
 
 read()
10 {
11     int x=0,y=1;char c=getchar();
12     while(c<‘0‘||c>‘9‘){if(c==‘-‘)y=-1;c=getchar();}
13     while(c>=‘0‘&&c<=‘9‘){x=(x<<3)+(x<<1)+c-‘0‘;c=getchar();}
14     return x*y;
15 }
16 ll T,L,d,n,m,ans,c;
17 bool flag=0;
18 ll gcd(ll x,ll y){return y==0
 
?x:gcd(y,x%y);}
19 ll phi(ll x)
20 {
21     ll sm=x,y=x;
22     go(i,2,sqrt(y))
23     {
24         if(x%i==0)sm=sm/i*(i-1);
25         while(x%i==0)x/=i;
26     }
27     if(x>1)sm=sm/x*(x-1);
28     return sm;
29 }
30 ll mul(ll x,ll y,ll mod)
31 {
32     ll sm=0;
33     while(y)
34     {
 
35         if(y&1)sm=(sm+x)%mod;
36         x=(x<<1)%mod;y>>=1;
37     }
38     return sm;
39 }
40 ll ksm(ll x,ll y,ll mod)
41 {
42     ll sm=1;
43     while(y)
44     {
45         if(y&1)sm=mul(sm,x,mod);//註意這裏直接乘起來會溢出
46         x=mul(x,x,mod);y>>=1;
47     }
48     return sm;
49 }
50 int main()
51 {
52     while(1)
53     {
54         scanf("%lld",&L);if(!L)break;T++;flag=0;ans=0;
55         if(8%L==0){printf("Case %lld: 1\n",T);continue;}
56         d=gcd(L,8);n=9*L/d;m=phi(n);c=sqrt(m);
57         go(i,2,c)
58         {
59             if(m%i==0&&ksm(10,i,n)==1){ans=i;flag=1;break;}
60         }
61         if(!flag)
62             yes(i,c,2)
63             {
64                 if(m%i==0&&ksm(10,m/i,n)==1){ans=m/i;break;}
65             }
66         printf("Case %lld: %lld\n",T,ans);
67     }
68     return 0;
69 }

Poj3696 The Lukiest Number