poj 3070 <矩阵快速幂【模板】求Fibonacci数列>

Fibonacci

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

09 999999999 1000000000 -1

Sample Output

034 626 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​​

通过这个题学会了快速求Fibonacci数列的方法

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node{
    int F[2][2];
};
int main()
{
    int n;
    while (scanf("%d",&n),n!=-1)
    {
        if (n==0)
        {
            printf("0\n");
            continue;
        }
        node lp,yi,bian,shen;
        lp.F[0][0]=lp.F[1][0]=lp.F[0][1]=1;lp.F[1][1]=0;
        yi.F[0][0]=yi.F[1][1]=1;yi.F[0][1]=yi.F[1][0]=0;
        while (n)
        {
            if (n%2)
            {
                shen.F[0][0]=(yi.F[0][0]*lp.F[0][0]+yi.F[0][1]*lp.F[1][0])%10000;
                shen.F[0][1]=(yi.F[0][0]*lp.F[0][1]+yi.F[0][1]*lp.F[1][1])%10000;
                shen.F[1][0]=(yi.F[1][0]*lp.F[0][0]+yi.F[1][1]*lp.F[1][0])%10000;
                shen.F[1][1]=(yi.F[1][0]*lp.F[0][1]+yi.F[1][1]*lp.F[1][1])%10000;
                yi=shen;
            }
            {
                bian.F[0][0]=(lp.F[0][0]*lp.F[0][0]+lp.F[0][1]*lp.F[1][0])%10000;
                bian.F[0][1]=(lp.F[0][0]*lp.F[0][1]+lp.F[0][1]*lp.F[1][1])%10000;
                bian.F[1][0]=(lp.F[1][0]*lp.F[0][0]+lp.F[1][1]*lp.F[1][0])%10000;
                bian.F[1][1]=(lp.F[1][0]*lp.F[0][1]+lp.F[1][1]*lp.F[1][1])%10000;
                lp=bian;
            }
            n/=2;
        }
        printf("%d\n",yi.F[0][1]);
    }
    return 0;
}