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题目：
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., printFn 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:
.
这个题目和CSU上面的一个题目差不多，点击打开我的博客
代码：
#include<iostream> using namespace std; int a, b, c, d; int x1, x2, x3, x4; void f(int n) { if (n == 0) { a = 1; b = 0; c = 0; d = 1; return; } f(n / 2); x1 = a*a + b*c; x2 = (a + d)*b; x3 = (a + d)*c; x4 = b*c + d*d; a = x1 % 10000; b = x2 % 10000; c = x3 % 10000; d = x4 % 10000; if (n % 2) { c += a; a = c - a; d += b; b = d - b; } } int main() { int n; while (cin >> n) { if (n < 0)break; f(n); cout << b << endl; } return 0; }
同样的，这个题目我也考虑了周期的问题。
因为这个题目是mod 10000，不是1000000007
一方面10000很小，另外一方面1000000007是素数，差别很大。
经过枚举，我找到了，斐波那契数列的第15000项是0，第15001项是1
刚好可以对应上第0项是0，第1项是1。
那么15000是不是最小周期呢？
（不得不说15000这个数真的很巧，1000000007那个周期是1000000007*2，
不知道是不是有某种数量关系）
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原文地址：http://blog.csdn.net/nameofcsdn/article/details/52254528
