POJ 3070 Fibonacci【矩阵快速幂取模】【模板题】

本文链接：https://blog.youkuaiyun.com/hurmishine/article/details/52986794

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13728		Accepted: 9724

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

Sample Output

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

原题链接：http://poj.org/problem?id=3070

递推的式子一般都可以化成矩阵相乘的形式。关键的就是构造矩阵，

但这题题目已经构造好了，所以就成了模（shui）板（shu）题了。

AC代码：

/**
    blog:http://blog.youkuaiyun.com/hurmishine
*/

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
const int mod= 10000;
LL n;
struct Mat
{
    int m[2][2];
};
Mat P=
{
    1,1,
    1,0
};
Mat I=
{
    1,0,
    0,1
};
Mat mul (Mat x,Mat y)
{
    Mat z;
    for(int i=0; i<2; i++)
    {
        for(int j=0; j<2; j++)
        {
            z.m[i][j]=0;
            for(int k=0; k<2; k++)
            {
                z.m[i][j]+=(x.m[i][k]*y.m[k][j])%mod;
                z.m[i][j]%=mod;
            }
        }
    }
    return z;
}
Mat quick_Mod(Mat a,LL b)
{
    Mat ans=I;
    Mat tmp=a;
    while(b>0)
    {
        if(b&1)
            ans=mul(ans,tmp);
        tmp=mul(tmp,tmp);
        b>>=1;
    }
    return ans;
}
int main()
{
    while(cin>>n)
    {
        if(n==-1) break;
        Mat tmp=quick_Mod(P,n);
        cout<<tmp.m[0][1]<<endl;
    }
    return 0;
}

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