POJ 3070(Fibonacci-矩阵幂)

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内容目录
Language:
Fibonacci
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 6733   Accepted: 4765

Description

 Fibonacci integer sequence指 F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. eg:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

Fibonacci 的矩阵乘法求法如下

.

Given an integer n, 求 Fn mod 10000.

Input

多组数据,每行一个 n (where 0 ≤ n ≤ 1,000,000,000). 数据以 −1 结尾.

Output

对每组数据打印一行 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

正宗矩阵幂练手题。

做了这题就差不多理解矩阵乘法了。

补充：递归一般能用矩阵乘法，如f[i]=g(f[i-1],f[i-2])...

    但是规划不行，如f[i]=max(....)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<functional>
using namespace std;
#define MAXN (1000000000)
#define F (10000)
struct M
{
	int a[3][3];
	M(int i){a[1][1]=a[1][2]=a[2][1]=1;a[2][2]=0;	}
	M(){memset(a,0,sizeof(a));	}
	friend M operator*(M a,M b)
	{
		M c;
		for (int i=1;i<=2;i++)
			for (int j=1;j<=2;j++)
				for (int k=1;k<=2;k++)
				{
					c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%F;
				}
		return c;
	}
	friend M pow(M a,int b)
	{
		if (b==1)
		{
			M c(1);
			return c;
		}
		else
		{
			M c=pow(a,b/2);
			c=c*c;
			if (b%2) return c*a;
			return c;
		}
	}
};
int n;
int main()
{
	while (cin>>n&&n!=-1)
	{
		if (n==0) printf("0n");
		else
		{
			M a=pow(M(1),n);
			printf("%dn",a.a[1][2]);
		}
	}
}