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:
.
解题思路：题目已经帮我们构造出矩阵行列式的关系了，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，很自然就想到了矩阵快速幂，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Fn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 struct Matrix{int m[maxn][maxn];}init;int n; 7 Matrix mul(Matrix a,Matrix b){//矩阵相乘 8 Matrix c; 9 for(int i=0;i<maxn;i++){//第一个矩阵的行 10 for(int j=0;j<maxn;j++){//第二个矩阵的列 11 c.m[i][j]=0; 12 for(int k=0;k<maxn;k++) 13 c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod; 14 } 15 } 16 return c; 17 } 18 Matrix POW(Matrix a,int x){ 19 Matrix b;memset(b.m,0,sizeof(b.m)); 20 for(int i=0;i<maxn;++i)b.m[i][i]=1;//单位矩阵 21 while(x){ 22 if(x&1)b=mul(b,a); 23 a=mul(a,a); 24 x>>=1; 25 } 26 return b; 27 } 28 int main(){ 29 while(cin>>n&&(n!=-1)){ 30 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 31 Matrix res = POW(init,n);//矩阵快速幂 32 cout<<res.m[0][1]<<endl;//取右上角的值即可 33 } 34 return 0; 35 }