本文主要是介绍[ACM] POJ 3070 Fibonacci （矩阵幂运算），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9517 | Accepted: 6767 |
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., print Fn 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
题目中给了提示矩阵。
网赛时才接触矩阵幂运算，自己掌握的知识还是太少了。。。。
代码：
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn=2;
const int mod=10000;
int n=2;//矩阵大小
int num;//幂大小struct mat
{int arr[maxn][maxn];mat(){memset(arr,0,sizeof(arr));}
};mat mul(mat a,mat b)
{mat ans;for(int i=0;i<n;i++){for(int k=0;k<n;k++){if(a.arr[i][k])for(int j=0;j<n;j++){ans.arr[i][j]+=a.arr[i][k]*b.arr[k][j];if(ans.arr[i][j]>=mod)ans.arr[i][j]%=mod;}}}return ans;
}mat power(mat p,int k)
{if(k==1) return p;mat e;for(int i=0;i<n;i++)e.arr[i][i]=1;if(k==0) return e;while(k){if(k&1)e=mul(e,p);p=mul(p,p);k>>=1;}return e;
}int main()
{mat ori;//题目中的原始矩阵ori.arr[0][0]=ori.arr[0][1]=ori.arr[1][0]=1;while(cin>>num&&num!=-1){mat ans;ans=power(ori,num);cout<<ans.arr[0][1]<<endl;}return 0;
}
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