POJ3070 Fibonacci （矩阵连乘）

                                                                                           Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3120		Accepted: 2203

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:

 

没办法  照片不能复制。这个题目会死我们大学的一个神牛给我做的，开始一头雾水，后来拨开迷雾，‘见得云月。说白了就是矩阵连乘加上二分。贴上代码：

#include #include using namespace std; int a[5],b[5],c[5],v[5],n; int main() { while(cin>>n&&n!=-1) { a[1]=1;a[2]=1;a[3]=1;a[4]=0; b[1]=1;b[2]=0;b[3]=0;b[4]=1; if(n==0) {cout<<0<>=1; } cout<