POJ 3070 Fibonacci (矩阵快速幂)

Fibonacci
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 10440   Accepted: 7421

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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:

.

解题思路：

　　关键是会求解转移矩阵，剩下的就是扔模板了。。

 

代码：

 1 # include<cstdio>
 2 # include<iostream>
 3 # include<fstream>
 4 
 5 using namespace std;
 6 
 7 # define MOD 10000
 8 
 9 typedef long long LL;
10 
11 struct matrix
12 {
13     LL a[2][2];
14     void init()
15     {
16         a[0][0] = 1;
17         a[0][1] = 1;
18         a[1][0] = 1;
19         a[1][1] = 0;
20     }
21 };
22 
23 int n;
24 
25 matrix fun ( matrix aa, matrix bb )
26 {
27     matrix cc;
28     for ( int i = 0;i < 2;i ++ )
29     {
30         for ( int j = 0;j < 2;j++ )
31         {
32             cc.a[i][j] = 0;
33             for ( int k = 0;k < 2;k++ )
34             {
35                 cc.a[i][j]+=(aa.a[i][k]*bb.a[k][j]);
36             }
37             cc.a[i][j]%=MOD;
38         }
39     }
40 
41     return cc;
42 }
43 
44 
45 
46 matrix My_power( matrix aa,int k )
47 {
48     matrix  ans;
49     ans.init();
50     while ( k >= 1 )
51     {
52         if ( k&1 )
53         {
54             ans = fun(ans,aa);
55         }
56         k/=2;
57         aa = fun(aa,aa);
58     }
59 
60     return ans;
61 
62 }
63 
64 
65 int main(void)
66 {
67 
68     int n;
69     while ( scanf("%d",&n)!=EOF )
70     {
71         if ( n==-1 )
72             break;
73         if ( n==0 )
74         {
75             printf("0\n");
76             continue;
77         }
78         matrix aa;
79         aa.init();
80         aa = My_power(aa,n-1);
81         printf("%lld\n",aa.a[0][1]%MOD);
82     }
83 
84     return 0;
85 }

 

posted @ 2015-05-17 00:17  BYYB_0506  阅读(137)  评论(0)  编辑  收藏  举报