（矩阵快速幂） Fibonacci -- poj -- 3070

链接：

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

Fibonacci

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:

.

代码：

#include<stdio.h>
#include<string.h>
#define MOD 10000
struct node
{
    int m[2][2];
}a, b;

node cheng(node x, node y)
{
    int i, j, k;
    node c;

    for(i=0; i<2; i++)
    for(j=0; j<2; j++)
    {
        c.m[i][j] = 0;
        for(k=0; k<2; k++)
            c.m[i][j] = (c.m[i][j] + x.m[i][k]*y.m[k][j])%MOD;
    }

    return c;
}

int Fast_MOD(int n)
{
    a.m[0][0] = a.m[0][1] = a.m[1][0] = 1;
    a.m[1][1] = 0;

    b.m[0][0] = b.m[1][1] = 1;  /// b 初始化为单位矩阵
    b.m[0][1] = b.m[1][0] = 0;

    while(n)
    {
        if(n&1) /// n是奇数
            b = cheng(b, a);
        a = cheng(a, a);
        n >>= 1;
    }
    return b.m[0][1];
}

int main()
{
    int n;
    while(scanf("%d", &n), n!=-1)
    {
        printf("%d\n", Fast_MOD(n));
    }
    return 0;
}