永夜初晗凝碧天

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poj3070来体会矩阵的妙用

Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5039 Accepted: 3479

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:

.

题目大意就是求出斐波那契数列第k项的后四位，如果有前导0就不输出0，但最后一个0必须输出

刚开始的时候就想到了一个关于斐波那契数列的数列的一个公式，准备用那个公式求的，但题目已经提示了用矩阵，就没用那个公式

有兴趣的可以看看这个公式，试试能不能那个解这题

a_{n}=\frac{\sqrt{5}}{5} \cdot \left[\left(\frac{1 + \sqrt{5}}{2}\right)^{n} - \left(\frac{1 - \sqrt{5}}{2}\right)^{n}\right]

这一题就要注意到矩阵

的右上角或左下角对应的值正好是第n项的值

故可以通过快速幂来求得斐波那契数列的哦n项，由于数据较大，还要记得要不断的对10000取模

矩阵的快速幂我是直接抄的模板，刚开始时不理解，现在懂了，详情直接看代码

http://www.cnblogs.com/ACShiryu

参考代码;

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 using namespace std;
 8 struct prog {
 9     int a[2][2] ;
10     void init(){
11         a[0][0]=a[1][0]=a[0][1]=1;
12         a[1][1]=0;
13     }
14 };
15 
16 prog matrixmul ( prog a ,prog b )
17 {
18     int i , j , k ;
19     prog c ;
20     for ( i = 0 ; i < 2; i ++ )
21     {
22         for ( j = 0 ; j < 2 ; j ++ )
23         {
24             c.a[i][j]=0;
25             for ( k =0 ; k < 2; k ++ )
26                 c.a[i][j]+=(a.a[i][k]*b.a[k][j]) ;
27             c.a[i][j] %= 10000 ;
28         }
29     }
30     return c ;
31 }
32 prog mul (prog s , int k )
33 {
34     prog ans ;
35     ans.init();
36     while ( k >= 1 )
37     {
38         if ( k & 1 )
39             ans = matrixmul ( ans , s ) ;
40         k = k >> 1 ;
41         s = matrixmul ( s , s ) ;
42     }
43     return ans ;
44 }
45 int main()
46 {
47     int n ;
48     while ( cin >> n , ~ n )
49     {
50         if ( ! n )
51         {
52             cout<<0<<endl ;
53             continue ;
54         }
55         prog s ;
56         s.init ( ) ;
57         s = mul ( s , n - 1 ) ;
58         cout << s . a [0][1] % 10000 <<endl;
59     }
60 
61     return 0;
62 }

posted on 2011-08-09 11:11  ACShiryu  阅读(1591)  评论(1)  编辑  收藏  举报