hud5451_求循环节加矩阵快速幂

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <ctime>
 8 #include <queue>
 9 #include <list>
10 #include <set>
11 #include <map>
12 using namespace std;
13 #define INF 0x3f3f3f3f
14 typedef long long LL;
15 
16 LL mod;
17 struct Mat
18 {
19     LL a[3][3];
20 };
21 LL pow(LL base, LL n, LL mod)
22 {
23     LL res = 1;
24     while(n)
25     {
26         if(n % 2)
27             res = (res * base) % mod;
28         base = (base * base) % mod;
29         n >>= 1;
30     }
31     return res;
32 }
33 Mat mul(Mat x, Mat y)
34 {
35     Mat z;
36     memset(z.a, 0, sizeof(z.a));
37     for(int i = 0; i < 2; i++)
38     {
39         for(int k = 0; k < 2; k++)
40         {
41             for(int j = 0; j < 2; j++)
42             {
43                 z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % mod) % mod;
44             }
45         }
46     }
47     return z;
48 }
49 void solve(LL n)
50 {
51     Mat res, base;
52     memset(res.a, 0, sizeof(res.a));
53     res.a[0][0] = 1, res.a[1][1] = 1;
54     base.a[0][0] = 5, base.a[0][1] = 2;
55     base.a[1][0] = 12, base.a[1][1] = 5;
56     while(n)
57     {
58         if(n % 2)
59             res = mul(res, base);
60         base = mul(base, base);
61         n >>= 1;
62     }
63     LL ans = 0;
64     ans = (ans + (res.a[0][0] * 5) % mod) % mod;
65     ans = (ans + (res.a[1][0] * 2) % mod) % mod;
66     ans = (ans * 2 - 1) % mod;
67     printf("%lld\n", ans);
68 }
69 
70 int main()
71 {
72     int t;
73     LL x;
74     scanf("%d", &t);
75     for(int i = 1; i <= t; i++)
76     {
77         scanf("%lld %lld", &x, &mod);
78         LL y = (mod + 1) * (mod - 1);
79         LL N = pow(2, x, y);
80         printf("Case #%d: ", i);
81         solve(N);
82     }
83     return 0;
84 }