矩阵快速幂 求斐波拉切数列的第n项 poj3070

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7241		Accepted: 5131

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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

求斐波拉切数列的第n项数取余后的值  但是问题在于n给的特别大

这时就要用到矩阵快速幂来解决这个问题，直接矩阵快速幂模板，其实和快速幂差不多只是把数换为矩阵即可，在遇到像斐波拉契这种有递推关系时可以考虑一下矩阵快速幂

难点就在于如何构建矩阵，就像上面这个题的矩阵就特别好构造

代码：

#include <cstdio>

#include <iostream>

using namespace std;

const int MOD = 10000;

struct matrix

{

    int m[2][2];

}ans, base;

matrix multi(matrix a, matrix b)

{

    matrix tmp;

    for(int i = 0; i < 2; ++i)

    {

        for(int j = 0; j < 2; ++j)

        {

            tmp.m[i][j] = 0;

            for(int k = 0; k < 2; ++k)

                tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;

        }

    }

    return tmp;

}

int fast_mod(int n)  // 求矩阵 base 的  n 次幂 

{

    base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;

    base.m[1][1] = 0;

    ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化为单位矩阵 

    ans.m[0][1] = ans.m[1][0] = 0;

    while(n)

    {

        if(n & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t 

        {

            ans = multi(ans, base);

        }

        base = multi(base, base);

        n >>= 1;

    }

    return ans.m[0][1];

}

int main()

{

    int n;

    while(scanf("%d", &n) && n != -1)

    {   

        printf("%d\n", fast_mod(n));

    }

    return 0;

}