poj 3070 Fibonacci 矩阵快速幂

题目链接：http://poj.org/problem?id=3070

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).

题意描述：菲波那契数列可以用题中那个矩阵计算得到，给出n（n可以超大），求出Fn。

算法分析：赤果果的矩阵快速幂。如果对矩阵快速幂不熟悉的话，可以先百度看看讲解。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
const int mod=10000;

int n;
struct matrix
{
    int an[2][2];
}res;
matrix multiply(matrix a,matrix b)
{
    matrix s;
    memset(s.an,0,sizeof(s.an));
    for (int i=0 ;i<2 ;i++)
    {
        for (int j=0 ;j<2 ;j++)
        {
            for (int k=0 ;k<2 ;k++)
            {
                s.an[i][j] += a.an[i][k]*b.an[k][j];
                if (s.an[i][j]>mod) s.an[i][j]%=mod;
            }
        }
    }
    return s;
}
void calc(int n)
{
    matrix sum;
    for (int i=0 ;i<2 ;i++)
    {
        for (int j=0 ;j<2 ;j++)
        sum.an[i][j]= i==j ? 1 : 0 ;
    }
    for (int i=0 ;i<2 ;i++)
    {
        for (int j=0 ;j<2 ;j++)
        res.an[i][j]=1;
    }
    res.an[1][1]=0;
    while (n)
    {
        if (n&1) sum=multiply(sum,res);
        n >>= 1;
        res=multiply(res,res);
    }
    printf("%d
",sum.an[1][0]%mod);
}

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