斐波那契数列（非水）

非水斐波那契数列

题目描述

斐波那契数列:1,1,2,3,5,8,13……请你编程计算出第n项%1000000007的值(n<=10^15).

输入

一个整数n（n<=10^15）

输出

数列第n项的值

样例输入

样例输出

提示

温馨提示:10^15>2^31-1

代码:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<vector>
 4 using namespace std; 
 5 
 6 const int Maxn = 1000000007; 
 7 
 8 struct matrix {
 9     long long m[2][2]; 
10 };
11 
12 matrix multi(matrix a, matrix b) {
13     long long sums = 0; 
14     matrix c; 
15     memset(c.m, 0, sizeof(c.m)); 
16     for (int i = 0; i <= 1; i++) {
17         for (int j = 0; j <= 1; j++) {
18             sums = 0; 
19             for (int k = 0; k <= 1; k++) {
20                 sums = (sums + a.m[i][k] * b.m[k][j]) % Maxn; 
21             }
22             c.m[i][j] = sums % Maxn; 
23         }
24     }
25     return c; 
26 }
27 
28 matrix matmod(matrix a, long long k) {
29     matrix res; 
30     memset(res.m, 0, sizeof(res.m)); 
31     for (int i = 0; i <= 1; i++) {
32         res.m[i][i] = 1; 
33     }
34     while(k) {
35         if (k & 1) {
36             res = multi(res, a); 
37         }
38         k = (k >> 1); 
39         a = multi(a, a); 
40     }
41     return res; 
42 }
43 
44 int main() {
45     long long n; 
46     scanf("%lld", &n); 
47     if (n < 3) {
48         printf("1"); 
49     }
50     else {
51         matrix a, b; 
52         memset(a.m, 0, sizeof(a.m)); 
53         memset(b.m, 0, sizeof(b.m)); 
54         for (int i = 0; i <= 1; i++) {
55             a.m[i][1] = 1; 
56         }
57         for (int i = 0; i <= 1; i++) {
58             b.m[i][0] = 1; 
59         }
60         a.m[1][0] = 1; 
61         a = matmod(a, n - 2); 
62         a = multi(a, b); 
63         printf("%d", a.m[1][0]);
64     }
65     return 0; 
66 }