Time Limit: 1000MS
Memory Limit: 65536K
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.
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.
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).
0
9
999999999
1000000000
-1
0
34
626
6875
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:
.
lz的第一道矩阵快速幂。
之前lz只知道快速幂,没想到还有矩阵快速幂。
那在这里就先稍微讲一下快速幂。
比如求x^11
11的二进制是1011,我们都知道(X^m)*(X^n) = X^(m+n)。所以同样我们也可以将x^11这样来做。(x^11 = x^8*x^2*x^1 = x^(8+2+1))
求x^n还有一种求法就是二分递归来求。利用x^n = x^(n/2)*x^(n/2) (n为偶数时),x^n = x^(n/2)*x^(n/2)*x(n为奇数时);
具体实现如下
这里再说一下
单位矩阵的定义: n*n的矩阵 Matrix ( i , i )=1; 任何一个矩阵乘以单位矩阵就是它本身 n*单位矩阵=n, 可以把单位矩阵等价为整数1。(单位矩阵用在矩阵快速幂中)
好,稍微了解一下后进人正题。
题意:求斐波那契数列的第n项%10000。
因为数据较大,而且题里给出了斐波那契的矩阵公式
这不就是求这个矩阵A(假设为A)的n次嘛。所以Fn就可以通过A^n来求出。
好,接下来上AC code(^_^)