矩阵快速幂 求Fibonacci数列poj3070

原作者博客地址：http://www.cnblogs.com/dongsheng/archive/2013/06/02/3114073.html

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

099999999991000000000-1

Sample Output

0346266875

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

分析：通过这道题，不仅学会了矩阵的快速幂的做法，同时也提供了求Fibonacci的高效算法

#include <cstdio> #include <iostream>  using namespace std;  const int MOD = 10000;  struct matrix {     int m[2][2]; }ans, base;  matrix multi(matrix a, matrix b) {     matrix tmp;     for(int i = 0; i < 2; ++i)     {         for(int j = 0; j < 2; ++j)         {             tmp.m[i][j] = 0;             for(int k = 0; k < 2; ++k)                 tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;         }     }     return tmp; } int fast_mod(int n)  // 求矩阵 base 的  n 次幂  {     base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;     base.m[1][1] = 0;     ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化为单位矩阵      ans.m[0][1] = ans.m[1][0] = 0;     while(n)     {         if(n & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t          {             ans = multi(ans, base);         }         base = multi(base, base);         n >>= 1;     }     return ans.m[0][1]; }  int main() {     int n;     while(scanf("%d", &n) && n != -1)     {            printf("%d\n", fast_mod(n));     }     return 0; }