nyoj fibonacci数列(二)行列乗算

http://acm.nyist.net/JudgeOnline/problem.php?pid=148
fibonacci数列(二)
時間制限:
1000 ms|メモリ制限:
65535 KB
難易度:
3
説明
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.
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:
.
入力
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.
しゅつりょく
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).
サンプル入力

0
9
1000000000
-1

サンプル出力

0
34
6875

その年に作ったこの~~はほとんど差がないことを忘れました~~大行列の乗算~~再帰的に解くことで得られます.

#include<stdio.h>
struct node{
           int a11,a12,a21,a22;
           }num,a;
node matrix(node a,node b)   //      
{
    node c;
    c.a11= (a.a11*b.a11%10000  + a.a12*b.a21%10000)%10000;
    c.a12= (a.a11*b.a12%10000 + a.a12*b.a22%10000)%10000;
    c.a21= (a.a21*b.a11%10000 + a.a22*b.a21%10000)%10000;
    c.a22= (a.a21*b.a12%10000 + a.a22*b.a22%10000)%10000;
    return c;     
}
node fun(int n)
{    
     if(n==1)return a;
     node c=fun(n/2);
     c=matrix(c,c);
     if(n%2==1)c=matrix(c,a);
     return c;
}



int main()
{

    a.a11=1,a.a12=1,a.a21=1,a.a22=0;    
   int n;
   node c;
   while(scanf("%d",&n) && n!=-1) 
   {
       if( n==0){printf("0
");continue;} 
       if( n==1){printf("1
");continue;}                     
       c=fun(n);     
       printf("%d
",c.a21);
   } 
    
}