矩阵快速幂 求斐波拉切数列的第n项poj3070

Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7241   Accepted: 5131

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:

.

Source

Stanford Local 2006

求斐波拉切数列的第n项数取余后的值  但是问题在于n给的特别大

这时就要用到矩阵快速幂来解决这个问题，直接矩阵快速幂模板，其实和快速幂差不多只是把数换为矩阵即可，在遇到像斐波拉契这种有递推关系时可以考虑一下矩阵快速幂

难点就在于如何构建矩阵，就像上面这个题的矩阵就特别好构造

代码：

#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; }