uva 10692 Huge Mods (欧拉函数，指数循环节)
题意：求一个楼梯数%m的大小。
思路：指数循环节。
需要注意的是，模数只有最外层是m，每往里一层，模数都变成m=phi(m)
所以可以写个dfs或者先预处理出每一层m存一下。
记得考虑n=1的特殊情况。
1/* ***********************************************
2Author :111qqz
3Created Time :Wed 26 Oct 2016 07:07:27 PM CST
4File Name :code/uva/10692.cpp
5************************************************ */
6#include <cstdio>
7#include <cstring>
8#include <iostream>
9#include <algorithm>
10#include <vector>
11#include <queue>
12#include <set>
13#include <map>
14#include <string>
15#include <cmath>
16#include <cstdlib>
17#include <ctime>
18#define fst first
19#define sec second
20#define lson l,m,rt<<1
21#define rson m+1,r,rt<<1|1
22#define ms(a,x) memset(a,x,sizeof(a))
23typedef long long LL;
24#define pi pair < int ,int >
25#define MP make_pair
26using namespace std;
27const double eps = 1E-8;
28const int dx4[4]={1,0,0,-1};
29const int dy4[4]={0,-1,1,0};
30const int inf = 0x3f3f3f3f;
31char st[20];
32LL n,m;
33LL a[15];
34LL ksm( LL a,LL b,LL k)
35{
36 LL res = 1;
37 while (b>0)
38 {
39 if (b&1) res = (res * a )% k;
40 b = b >> 1;
41 a = ( a * a) % k;
42 }
43 return res;
44}
45LL euler( LL x)
46{
47 LL ret = 1 ;
48 for ( LL i = 2 ; i*i <= x ; i++)
49 {
50 if (x%i==0)
51 {
52 x/=i;
53 ret*=(i-1);
54 while (x%i==0)
55 {
56 x/=i;
57 ret*=i;
58 }
59 }
60 }
61 if (x>1) ret*=(x-1);
62 return ret;
63}
64LL phi[20];
65int main()
66{
67 #ifndef ONLINE_JUDGE
68 freopen("code/in.txt","r",stdin);
69 #endif
70 int cas = 0 ;
71 while (~scanf("%s",st))
72 {
73 ms(a,0);
74 ms(phi,0);
75 // cout<<"st:"<<st<<endl;
76 if (st[0]=='#') break;
77 int len = strlen(st);
78 m = 0 ;
79 for ( int i = 0 ; i < len ; i++)
80 {
81 LL val = st[i]-'0';
82 m = m * 10 + val;
83 }
84 // cout<<"m:"<<m<<endl;
85 scanf("%lld",&n);
86 phi[0] = m;
87 for ( int i = 1; i <= n ; i++) scanf("%lld",a+i);
88 LL fi = euler(m);
89 for ( int i = 1 ; i <= n-1 ;i++)
90 {
91 phi[i] = fi;
92 fi = euler(fi);
93 }
94 for ( int i = n-1 ; i >=1 ; i--)
95 {
96 a[i] = ksm(a[i],(a[i+1]%phi[i]+phi[i]),phi[i-1]);
97 }
98 printf("Case #%d: %lld\n",++cas,a[1]%m); //考虑n为1的情况
99 ms(st,0);
100 }
101 #ifndef ONLINE_JUDGE
102 fclose(stdin);
103 #endif
104 return 0;
105}