题解报告：poj 3070 Fibonacci

题目链接：http://poj.org/problem?id=3070

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

Sample Output

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:

解题思路：题目已经说得清楚，这里就讲讲思路吧。通过给出的递推式我们可以推导得到题目那个n次幂的矩阵的通项公式，记那个矩阵为A，所以问题转化成求Aⁿ，然后就不自觉想到了矩阵快速幂，没错，无论n有多大，采用矩阵快速幂的做法就是一件很简单的事情。不过如果n值超过int最大值，要改成long long类型，避免数据溢出，其余代码不变。最后取结果矩阵右上角的值即为F_n对应的结果。

AC代码：

 1 #include<iostream>
 2 #include<string.h>
 3 using namespace std;
 4 const int mod=10000;
 5 const int maxn=2;//2行2列式
 6 int n;
 7 struct Matrix
 8 {
 9     int m[maxn][maxn];
10 }init;
11 Matrix mul(Matrix a,Matrix b)//矩阵相乘
12 {
13     Matrix c;
14     for(int i=0;i<maxn;i++){
15         for(int j=0;j<maxn;j++){
16             c.m[i][j]=0;
17             for(int k=0;k<maxn;k++)
18                 c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j]%mod)%mod;
19             c.m[i][j]%=mod;
20         }
21     }
22     return c;
23 }
24 Matrix POW(Matrix a,int x)
25 {
26     Matrix b;
27     memset(b.m,0,sizeof(b.m));
28     for(int i=0;i<maxn;++i)b.m[i][i]=1;//单位矩阵
29     while(x){
30         if(x&1)b=mul(b,a);
31         a=mul(a,a);
32         x>>=1;
33     }
34     return b;
35 }
36 int main()
37 {
38     while(cin>>n && (n!=-1)){
39         init.m[0][0]=1;init.m[0][1]=1;init.m[1][0]=1;init.m[1][1]=0;//初始化矩阵
40         Matrix res = POW(init,n);//矩阵快速幂
41         cout<<res.m[0][1]<<endl;//取右上角的值即可
42     }
43     return 0;
44 }

题解报告：poj 3070 Fibonacci

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