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！