Fibonacci（矩阵）

Fibonacci

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

 

Description

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.

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 F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n 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:

.

 

 

题意：求Fibonacci数列的第n项。

注意n很大，用矩阵，就可以了。

题目链接：http://poj.org/problem?id=3070  转载请注明出处：寻找&星空の孩子

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

#define LL __int64

#define mod 10000

struct matrix

{

    LL mat[2][2];

};



matrix multiply(matrix a,matrix b)

{

    matrix c;

    memset(c.mat,0,sizeof(c.mat));

    for(int i=0;i<2;i++)

    {

        for(int j=0;j<2;j++)

        {

            if(a.mat[i][j]==0)continue;

            for(int k=0;k<2;k++)

            {

                if(b.mat[j][k]==0)continue;

                c.mat[i][k]+=a.mat[i][j]*b.mat[j][k]%mod;

                c.mat[i][k]%=mod;

            }

        }

    }

    return c;

}



matrix quicklymod(matrix a,LL n)

{

    matrix res;

    memset(res.mat,0,sizeof(res.mat));

    for(int i=0;i<2;i++) res.mat[i][i]=1;

    while(n)

    {

        if(n&1)

            res=multiply(a,res);

        a=multiply(a,a);

        n>>=1;

    }

    return res;

}



int main()

{

    LL n;

    while(scanf("%I64d",&n)!=EOF)

    {

        if(n==-1)break;

        matrix ans;

        ans.mat[0][0]=1;

        ans.mat[0][1]=1;

        ans.mat[1][0]=1;

        ans.mat[1][1]=0;



        if(n==0)

        {

            printf("0\n");

            continue;

        }

 /*       else if(n==1)

        {

            printf("1\n");

            continue;

        }*/



        else

            ans=quicklymod(ans,n);



 /*       for(int i=0; i<2; i++)

        {

            for(int j=0; j<2; j++)

                printf("%I64d\t",ans.mat[i][j]);

            printf("\n");

        }

        printf("\n");

*/

        printf("%I64d\n",ans.mat[1][0]);

    }

    return 0;

}

矩阵入门题！