fibonacci数列(二)

1000 ms | 内存限制： 65535

3

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.

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:

. 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.

输出 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). 样例输入

091000000000-1

样例输出

0346875

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#define ll long long
#define M 10000
#define N 2
using namespace std;
struct mat
{
	ll m[N][N];
};
mat A=
{
	1,1,
	1,0
};
mat I=
{
	1,0,
	0,1
};
mat multi(mat a,mat b)
{
	mat c;
	for(int i=0;i<N;i++)
	{
		for(int j=0;j<N;j++)
		{
			c.m[i][j]=0;
			for(int k=0;k<N;k++)
				c.m[i][j]+=a.m[i][k]*b.m[k][j]%M;
			c.m[i][j]%=M;
		}
	}
	return c;
}
mat power(mat A,int k)//矩阵快速幂 
{
	mat ans=I,p=A;
	while(k)
	{
		if(k&1)
		{
			ans=multi(ans,p);
			k--;
		}
		k>>=1;
		p=multi(p,p);
	}
	return ans;
}
int main()
{
	int n;
	while(scanf("%d",&n)&&n!=-1)
	{
		if(n==0)
		{
			printf("0\n");
			continue;
		}
		mat ans=power(A,n-1);
		printf("%lld\n",ans.m[0][0]);
	}
	return 0;
}