矩阵快速幂求Fibonacci数列

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7241		Accepted: 5131

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

code:

 #include 
 2 #include 
 3 
 4 using namespace std;
 5 
 6 const int MOD = 10000;
 7 
 8 struct matrix
 9 {
10     int m[2][2];
11 }ans, base;
12 
13 matrix multi(matrix a, matrix b)
14 {
15     matrix tmp;
16     for(int i = 0; i < 2; ++i)
17     {
18         for(int j = 0; j < 2; ++j)
19         {
20             tmp.m[i][j] = 0;
21             for(int k = 0; k < 2; ++k)
22                 tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
23         }
24     }
25     return tmp;
26 }
27 int fast_mod(int n)  // 求矩阵 base 的  n 次幂 
28 {
29     base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;
30     base.m[1][1] = 0;
31     ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化为单位矩阵 
32     ans.m[0][1] = ans.m[1][0] = 0;
33     while(n)
34     {
35         if(n & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t 
36         {
37             ans = multi(ans, base);
38         }
39         base = multi(base, base);
40         n >>= 1;
41     }
42     return ans.m[0][1];
43 }
44 
45 int main()
46 {
47     int n;
48     while(scanf("%d", &n) && n != -1)
49     {   
50         printf("%d\n", fast_mod(n));
51     }
52     return 0;
53 }