POJ 3070 Fibonacci【斐波那契数列/矩阵快速幂】

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17171		Accepted: 11999

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:

.

Source

Stanford Local 2006

【分析】：矩乘其实很简单，通过自己构造或者是搜索对于一个递推公式求出它所对应的常数矩阵，然后套个快速幂就可以迅速求解第n项。最后输出的是矩阵最左上方的值。根据前面的一些思路，现在我们需要构造一个2 x 2的矩阵，使得它乘以(a,b)得到的结果是(b,a+b)。每多乘一次这个矩阵，这两个数就会多迭代一次。那么，我们把这个2 x 2的矩阵自乘n次，再乘以(0,1)就可以得到第n个Fibonacci数了。不用多想，这个2 x 2的矩阵很容易构造出来。

【代码】：

#include <iostream>
#include <cstddef>
#include <cstring>
#include <vector>

using namespace std;

typedef long long ll;
const int mod=10000;
typedef vector<ll> vec;
typedef vector <vec> mat;

mat mul(mat &a,mat &b)
{
    mat c(a.size(),vec(b[0].size()));
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
        for(int k=0;k<2;k++){
          c[i][j]+=a[i][k]*b[k][j];
          c[i][j]%=mod;
        }
        return c;
}

mat Pow(mat a,ll n)
{
    mat res(a.size(),vec(a.size()));
    for(int i=0;i<a.size();i++)
        res[i][i]=1;
    while(n)
    {
        if(n&1)
            res=mul(res,a);
        a=mul(a,a);
        n/=2;
    }
    return res;
}

ll solve(ll n)
{
    mat a(2,vec(2));
    a[0][0]=1;
    a[0][1]=1;
    a[1][0]=1;
    a[1][1]=0;
    a=Pow(a,n);
    return a[0][1];
}

int main()
{
    ll n;
    while(cin>>n&&n!=-1)
    {
        cout<<solve(n)<<endl;
    }
}

斐波那契快速幂