【原】 POJ 3070 Fibonacci 斐波那契数列 解题报告

 

http://poj.org/problem?id=3070

方法：
divide and conquer，类似于求幂
矩阵求幂
复杂度：O(logn)

| F(n+1) F(n)   | = | 1  1 |^n
| F(n)   F(n-1) |   | 1  0 |

 

注意：
矩阵的0次幂是单位矩阵

 

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

 

   1: #include <stdio.h>

   2:  

   3: struct Matrix

   4: {

   5:     Matrix( int a=1, int b=1, int c=1, int d=0 ):a11(a),a12(b),a21(c),a22(d){}

   6:     void set( int a=1, int b=1, int c=1, int d=0 ){a11=a;a12=b;a21=c;a22=d;}

   7:     int a11; int a12; int a21; int a22;

   8: };

   9:  

  10: Matrix MatrixMultiply( Matrix& a, Matrix& b )

  11: {

  12:     Matrix result ;

  13:     result.a11 = ( a.a11*b.a11 + a.a12*b.a21 ) % 10000 ;

  14:     result.a12 = ( a.a11*b.a12 + a.a12*b.a22 ) % 10000 ;

  15:     result.a21 = ( a.a21*b.a11 + a.a22*b.a21 ) % 10000 ;

  16:     result.a22 = ( a.a21*b.a12 + a.a22*b.a22 ) % 10000 ;

  17:  

  18:     return result ;

  19: }

  20:  

  21: Matrix Fibonacci( Matrix& a, int n )

  22: {

  23:     Matrix result ;

  24:     if( n==0 )

  25:     {

  26:         result.set(1,0,0,1) ;

  27:         return result ;

  28:     }

  29:     if( n==1 )

  30:         return result ;

  31:     

  32:     Matrix tmp ;

  33:     tmp = Fibonacci( a, n/2 ) ;

  34:     result = MatrixMultiply( tmp, tmp ) ;

  35:     if( n%2==0 )

  36:         return result;

  37:     else

  38:         return MatrixMultiply(a,result) ;

  39: }

  40:  

  41: void run3070()

  42: {

  43:     Matrix a,result ;

  44:     int n ;

  45:  

  46:     while( scanf( "%d", &n ) && n!=-1 )

  47:     {

  48:         result = Fibonacci( a, n ) ;

  49:         printf( "%d\n", result.a12 ) ;

  50:     }

  51: }