好玩的Fibonacci数列

1.原理：

不做过多解释，不会去翻高中数学教材

2.重要结论：

　　1.前n项和 可以 用 n + 2项的数减一得到

 

计算方法：

　　矩阵快速幂！！！

上板子题：

 

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21213		Accepted: 14511

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

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

#include <vector> #include <iostream>     using namespace std;   typedef long long ll; typedef vector<ll> vec; typedef vector<vec> mat;   const ll  mod = 10000;   mat mul(mat & A, mat& B) {     mat C(A.size(), vec(B[0].size()));     for (int i = 0; i < A.size(); ++i)         for (int k = 0; k < B.size(); ++k)             if (A[i][k])                 for (int j = 0; j < B[0].size(); ++j)                     C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;     return C;            }     mat Pow(mat A, ll n) {     mat B(A.size(), vec(A.size()));     for (int i = 0; i < A.size(); ++i)         B[i][i] = 1;     for (; n; n >>= 1, A = mul(A, A))         if(n & 1)             B = mul(B, A);     return B; }     int main (){     mat A(2);       A.resize(2);//r行     for (int k = 0; k < 2; ++k){         A[k].resize(2);//每行为c列     }       A[0][0] = 1;     A[0][1] = 1;     A[1][0] = 1;     A[1][1] = 0;         ll n;     while(cin >> n && n != -1) {         if (n) {             mat C = Pow(A, n - 1);             cout << C[0][0] << endl;         }         else             cout << 0 << endl;           } }