poj 3070 矩阵连乘简单题_poj 矩阵连乘-优快云博客

本文链接：https://blog.youkuaiyun.com/yu_ming_cong/article/details/47667035

本文介绍了一种高效计算斐波那契数列中指定项最后四位数字的方法。通过矩阵快速幂运算，解决了大规模数值计算的问题，并提供了一个C++实现示例。

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Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11005		Accepted: 7833

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, andF_n =F_n_{− 1} + F_n_{− 2} forn ≥ 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 ofF_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., printF_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
代码：

#include <iostream>

using namespace std;
const int mod=10000;
struct Matrix
{
    long long m[2][2];
};
Matrix p={1,1,
          1,0};
Matrix I={1,0,
          0,1};
Matrix qm(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0;i<2;i++)
     {
         for(int j=0;j<2;j++)
         {   c.m[i][j]=0;
             for(int k=0;k<2;k++)
             {
                 c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;
             }
             c.m[i][j]=c.m[i][j]%mod;
         }
     }
     return c;
}
Matrix qp(long long n)
{
    Matrix m=p,b=I;
    while(n>=1)
    {
        if(n&1) b=qm(b,m);
        n=n>>1;
        m=qm(m,m);
    }
    return b;
}
int main()
{

    long long n;
    while(cin>>n)
    {  if(n==-1) break;
        Matrix ans=qp(n);
        cout<<ans.m[1][0]<<endl;
    }
    return 0;
}