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
Problem 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.
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
Sample Output
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[n],表示n位全是8的数，那么f[n]=10*f[n-1]+8,那么构造等比数列f[n]+(8/9)=10*(f[n-1]+(8/9));
那么f[n] = (8+8/9)*(10)^(n-1)-8/9;f[n] = (8/9)*（(10)^n-1）;那么就是要求最小的n使f[n]%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；
1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<string.h>
5 #include<queue>
6 #include<set>
7 #include<math.h>
8 #include<map>
9 using namespace std;
10 typedef long long LL;
11 pair<LL,LL>exgcd(LL n,LL m);
12 LL gcd(LL n,LL m);
13 LL quick(LL n,LL m,LL mod);
14 LL mul(LL n, LL m,LL p);
15 int slove(LL n);
16 LL phi(LL n);
17 bool prime[1000005];
18 int ans[1000005];
19 LL fen[1000005];
20 int main(void)
21 {
22 LL n;
23 int i,j;
24 int cn = 0;
25 for(i = 2; i <= 1000; i++)
26 {
27 if(!prime[i])
28 {
29 for(j = i; (i*j) <= 1000000; j++)
30 {
31 prime[i*j] = true;
32 }
33 }
34 }
35 for(i = 2; i <= 1000000; i++)
36 {
37 if(!prime[i])
38 {
39 ans[cn++] = i;
40 }
41 }
42 //printf("%d\n",cn);
43 int __ca = 0;
44 while(scanf("%lld",&n),n!=0)
45 {
46 LL gc = gcd(8,n);
47 n = 9*n/gc;
48 LL oula = phi(n);
49 LL x = gcd(n,10);//printf("%lld\n",n);
50 //printf("%lld\n",x);
51 printf("Case %d: ",++__ca);
52 if(x!=1)
53 {
54 printf("0\n");
55 }
56 else
57 {
58 int k = slove(oula);
59 //printf("%d\n",k);
60 for(i = 0;i < k;i++)
61 {
62 LL akk =quick(10,fen[i],n);
63 if(akk==1)
64 {
65 break;
66 }
67 }//printf("%d\n",10);
68 printf("%lld\n",fen[i]);
69 }
70 }
71 return 0;
72 }
73 int slove(LL n)
74 { int cn = 0;int i,j;
75 for(i = 1;i < sqrt(1.0*n);i++)
76 {
77 if(n%i==0)
78 {
79 if(n/i==i)
80 {
81 fen[cn++] = i;
82 }
83 else
84 {
85 fen[cn++] = i;
86 fen[cn++] = n/i;
87 }
88 }
89 }
90 sort(fen,fen+cn);
91 return cn;
92 }
93 LL phi(LL n)
94 {
95 int f = 0;
96 bool flag = false;
97 LL ask =n;
98 while(n>1)
99 {
100 while(n%ans[f]==0)
101 {
102 if(!flag)
103 {
104 flag = true;
105 ask/=ans[f];
106 ask*=ans[f]-1;
107 }
108 n/=ans[f];
109 }
110 f++;
111 flag = false;
112 if((LL)ans[f]*(LL)ans[f]>n)
113 {
114 break;
115 }
116 }
117 if(n > 1)
118 {
119 ask/=n;
120 ask*=(n-1);
121 }
122 return ask;
123 }
124 pair<LL,LL>exgcd(LL n,LL m)
125 {
126 if(m==0)
127 return make_pair(1,0);
128 else
129 {
130 pair<LL,LL>ak = exgcd(m,n%m);
131 return make_pair(ak.second,ak.first-(n/m)*ak.second);
132 }
133 }
134 LL gcd(LL n,LL m)
135 {
136 if(m==0)
137 return n;
138 else return gcd(m,n%m);
139 }
140 LL quick(LL n,LL m,LL mod)
141 {
142 LL ak = 1;
143 n %= mod;
144 while(m)
145 {
146 if(m&1)
147 ak =mul(ak,n,mod);
148 n = mul(n,n,mod);
149 m>>=1;
150 }
151 return ak;
152 }
153 LL mul(LL n, LL m,LL p)
154 {
155 n%=p;
156 m%=p;
157 LL ret=0;
158 while(m)
159 {
160 if(m&1)
161 {
162 ret=ret+n;
163 ret%=p;
164 }
165 m>>=1;
166 n<<=1;
167 n%=p;
168 }
169 return ret;
170 }