poj 3070 Fibonacci（矩阵加速DP）

Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12459   Accepted: 8853 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 Stanford Local 2006

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题解：矩阵加速DP。

学到了一个很高效的求解fibonacci的方法。

 1 1                                                                     fn+1  fn       1  1      fn+1+fn     fn+1

 1 0    这个矩阵既是加速矩阵也是初始矩阵。   fn    fn-1   ×  1  0   =fn+fn+1     fn         

所以要求序列第K项的答案就是求矩阵A^k  的   a[1][2]  

 #include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define N 2
#define p 10000
using namespace std;
int n,m;
struct data 
{
	int a[10][10];
}a;
void clear(data &x)
{
	for (int i=1;i<=N;i++)
	 for (int j=1;j<=N;j++)
	  x.a[i][j]=0;
}
void change(data &a,data b)
{
	for (int i=1;i<=N;i++)
	 for (int j=1;j<=N;j++)
	  a.a[i][j]=b.a[i][j];
}
data mul(data a,data b)
{
	data c;
	for (int i=1;i<=N;i++)
	 for (int j=1;j<=N;j++)
	  {
	  	c.a[i][j]=0;
	  	for (int k=1;k<=N;k++)
	  	 c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j]%p)%p;
	  }
	return c;
}
data pow(data num,int x)
{
	data ans; clear(ans);
	for (int i=1;i<=N;i++) ans.a[i][i]=1;
	data base; 
	change(base,num);  
	while (x)
	{
		if (x&1)  ans=mul(ans,base);
		x>>=1;
		base=mul(base,base);
	}
	return ans;
}
int main()
{
	freopen("a.in","r",stdin);
	a.a[1][1]=1; a.a[1][2]=1; a.a[2][1]=1; a.a[2][2]=0;
	while (scanf("%d",&n)!=EOF)
	{
		if (n==-1) break;
		if (n==0){
			printf("0\n");
			continue;
		}
		data ans=pow(a,n);
		printf("%d\n",ans.a[1][2]);
	}
}