poj 3070 Fibonacci【矩阵斐波拉切】

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11566 Accepted: 8218

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:

搁置了很久的题目，因为自己不懂矩阵的性质，不知道乘了之后会发生什么，也就根本搞不懂矩阵快速幂的方法了............

先这样吧，暂时理解还不到位，暂时记住吧，慢慢理解！

#include<stdio.h>#include<string.h>#define M 10000struct matrix{int m[2][2];}base,ans;void init(){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.m[0][1]=ans.m[1][0]=0;}matrix mul(matrix a,matrix b){matrix tp={0,0,0,0};for(int i=0;i<2;++i)//这三层循环可以有很多种组合方法，这样的可以便于控制剪枝{for(int j=0;j<2;++j){if(a.m[i][j]==0){continue;}for(int k=0;k<2;++k){tp.m[i][k]=(tp.m[i][k]+a.m[i][j]*b.m[j][k])%M;}}}return tp;}int main(){int n;while(scanf("%d",&n),n!=-1){init();while(n){if(n&1){ans=mul(ans,base);}base=mul(base,base);n>>=1;}printf("%d\n",ans.m[0][1]);}return 0;}