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 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:

.

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：

# include<cstdio>
# include<iostream>
# include<fstream>

using namespace std;

# define MOD 10000

typedef long long LL;

struct matrix
{
    LL a[2][2];
    void init()
    {
        a[0][0] = 1;
        a[0][1] = 1;
        a[1][0] = 1;
        a[1][1] = 0;
    }
};

int n;

matrix fun ( matrix aa, matrix bb )
{
    matrix cc;
    for ( int i = 0;i < 2;i ++ )
    {
        for ( int j = 0;j < 2;j++ )
        {
            cc.a[i][j] = 0;
            for ( int k = 0;k < 2;k++ )
            {
                cc.a[i][j]+=(aa.a[i][k]*bb.a[k][j]);
            }
            cc.a[i][j]%=MOD;
        }
    }

    return cc;
}



matrix My_power( matrix aa,int k )
{
    matrix  ans;
    ans.init();
    while ( k >= 1 )
    {
        if ( k&1 )
        {
            ans = fun(ans,aa);
        }
        k/=2;
        aa = fun(aa,aa);
    }

    return ans;

}


int main(void)
{

    int n;
    while ( scanf("%d",&n)!=EOF )
    {
        if ( n==-1 )
            break;
        if ( n==0 )
        {
            printf("0
");
            continue;
        }
        matrix aa;
        aa.init();
        aa = My_power(aa,n-1);
        printf("%lld
",aa.a[0][1]%MOD);
    }

    return 0;
}