codevs1281 矩阵乘法 快速幂 !!!手写乘法取模!!! 练习struct的构造函数和成员函数

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 
 5 
 6 struct data {
 7     long long f[3][3];
 8     data (void) {
 9         memset(f, 0, sizeof(f));
10         f[0][0] = 1;
11         f[1][1] = 1;
12         f[2][2] = 1;
13     }
14     void clear(void) {
15         memset(f, 0, sizeof(f));
16     }
17     void print(void) {
18         for (int i = 0; i < 3; i++) {
19             for (int j = 0; j < 3; j++) printf("%lld ", f[i][j]);
20             printf("\n");
21         }
22         printf("\n");
23     }
24 };
25 
26 long long m, a, c, x0, n, g;
27 data cur;
28 
29 long long mu (long long a1, long long a2) {
30     long long ans = 0;
31     while (a2 > 0) {
32         if (a2 & 1) ans = (ans + a1) % m;
33         a1 = (a1 + a1) % m;
34         a2 >>= 1;
35     }
36     return (ans);
37 }
38 
39 data operator * (data a1, data a2) {
40     data t;
41     t.clear();
42     for (int i = 0; i < 3; i++)
43         for (int j = 0; j < 3; j++) 
44             for (int k = 0; k < 3; k++) {
45                 t.f[i][j] = (t.f[i][j] + mu(a1.f[i][k], a2.f[k][j]) % m) % m;
46             }
47     return (t);
48 }
49 
50 data qpow(data x, long long v) {
51     data ans;
52     while (v > 0) {
53         if (v & 1) ans = ans * x;
54         x = x * x;
55         v >>= 1;
56     }
57     return (ans);
58 }
59 
60 int main () {
61     scanf("%lld %lld %lld %lld %lld %lld", &m, &a, &c, &x0, &n, &g);
62     cur.clear();
63     cur.f[1][0] = 1;
64     cur.f[1][1] = a % m;
65     cur.f[2][1] = 1;
66     cur.f[2][2] = 1;
67     cur = qpow(cur, n);
68     //cur.print();
69     long long ans = (mu(x0,cur.f[1][1]) % m + mu(c, cur.f[2][1] % m)) % m; 
70     ans = ((ans % g) + g) % g;
71     printf("%lld", ans);
72     return 0;
73 }