fzu Problem 2198 快来快来数一数 (快速幂+优化)

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <queue>
 7 #include <cstring>
 8 using namespace std;
 9 
10 #define LL long long
11 const LL maxn = 3;
12 const LL mod = 1000000007;
13 struct mat
14 {
15     LL col, row;
16     LL p[maxn][maxn];
17 } pp[65];
18 
19 mat mul (mat a, mat b);
20 mat pow (LL n, LL res, mat b);
21 
22 int main ()
23 {
24     LL n, t;
25     mat b;
26     scanf ("%lld", &t);
27     memset (pp[0].p, 0, sizeof(pp[0].p));
28     memset (b.p, 0, sizeof(b.p));
29     pp[0].col = pp[0].row = b.row = maxn;
30     b.col = 1;
31     pp[0].p[0][0] = 6;
32     pp[0].p[1][0] = -1;
33     pp[0].p[0][1] = pp[0].p[2][0] = pp[0].p[2][2] = 1;
34     b.p[0][0] = 6;
35     b.p[0][1] = 0;
36     b.p[0][2] = 5;
37     for (int i=1; i<65; i++)
38         pp[i] = mul(pp[i-1], pp[i-1]);
39         
40     while (t --)
41     {
42         mat a;
43         scanf ("%lld", &n);
44         a = pow(n-1, 0, b);
45         printf ("%lld\n", a.p[0][0] % mod);
46     }
47     
48     return 0;
49 }
50 
51 mat mul (mat a, mat b)
52 {
53     mat c;
54     c.col = a.col;
55     c.row = b.row;
56     memset (c.p, 0, sizeof(c.p));
57     for (int k=0; k<a.row; k++)
58         for (int i=0; i<a.col; i++)
59         {
60             if (a.p[i][k] == 0) continue;
61             for (int j=0; j<b.row; j++)
62             {
63                 if (b.p[k][j] == 0) continue;
64                 c.p[i][j] = (c.p[i][j] + a.p[i][k] * b.p[k][j] + mod) % mod;
65             }
66         }
67     return c;
68 }
69 
70 mat pow (LL n, LL res, mat b)
71 {
72     while (n)
73     {
74         if (n % 2)
75             b = mul (b, pp[res]);
76         res ++;
77         n /= 2;
78     }
79     return b;
80 }