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快速Fibonacci数列,矩阵法

Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9156 Accepted: 6494

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:

.


这个题是要用矩阵的知识解决斐波那契数列,题目中给的公式很简单,直接往上套就行了,那我们就先看除了题目给的方法外的另一个方法,当然还是矩阵。

先看下面的公式:


不难看出这个式子是把:


写成矩阵的形式而已。

将上面的式子迭代可得:


故可得一般式为:


继续展开会得到:


设:


则:


所以:


现在已经很明显了,其中还要用到矩阵快速幂的知识,代码如下:

#include <stdio.h>
#include <string.h>
#include <math.h>
typedef __int64 int64;

int64 c[2][2],ans[2][2],d[2][2],a[2][2];

int main()
{
	int64 i,j,n,k,t;
	while(scanf("%I64d",&n)!=EOF)
	{
		if(n==-1)		
			break;
		if(n==0)
		{
			printf("0\n");
			continue;
		}
		if(n==1)
		{
			printf("1\n");
			continue;
		}
		memset(ans,0,sizeof(ans));
		for(i=0;i<2;i++)
			ans[i][i]=1;
		a[0][0]=a[0][1]=a[1][0]=1;
		a[1][1]=0;
		n=n-2;
		while(n!=0)
		{
			if(n%2==1)
			{
				memset(d,0,sizeof(d));					
				for(i=0;i<2;i++)
					for(j=0;j<2;j++)
						if(a[i][j])
							for(k=0;k<2;k++)
							{
								d[i][k]+=ans[i][j]*a[j][k];		
								d[i][k]=d[i][k]%10000;
							}
				for(i=0;i<2;i++)
					for(j=0;j<2;j++)
						ans[i][j]=d[i][j];				
			}
			memset(c,0,sizeof(c));
			for(i=0;i<2;i++)
					for(j=0;j<2;j++)
					{
						if(a[i][j]==0)
							continue;
						for(k=0;k<2;k++)
						{
							c[i][k]+=a[i][j]*a[j][k]; 
							c[i][k]=c[i][k]%10000;
						}
					}
			for(i=0;i<2;i++)
				for(j=0;j<2;j++)
					a[i][j]=c[i][j];
			n=n/2;
		}
		t=(ans[0][0]+ans[0][1])%10000;
		if(t==0)
			printf("0\n");
		else 
			printf("%I64d\n",t);
	}
	return 0;
}