poj 3070 Fibonacci（矩阵快速幂）

Fibonacci

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:

.

ps：连公式都给出来了，直接矩阵快速幂……

代码：

#include<stdio.h>#include<string.h>#define N 2#define mod 10000#define mem(a,b) memset(a,b,sizeof(a))typedef long long LL;struct Matrix{    LL mat[N][N];};Matrix unit_matrix={    1,0,    0,1};//单位矩阵Matrix mul(Matrix a,Matrix b)//矩阵相乘{    Matrix res;    for(int i=0; i<N; ++i)        for(int j=0; j<N; ++j)        {            res.mat[i][j]=0;            for(int k=0; k<N; ++k)                res.mat[i][j]=(res.mat[i][j]+a.mat[i][k]*b.mat[k][j])%mod;        }    return res;}Matrix pow_matrix(Matrix a,LL n)//矩阵快速幂{    Matrix res=unit_matrix;    while(n)    {        if(n&1)            res=mul(res,a);        a=mul(a,a);        n>>=1;    }    return res;}int main(){    LL n;    Matrix tmp,arr;    while(~scanf("%lld",&n),n!=-1)    {        if(n==0)            printf("0\n");        else if(n==1)            printf("1\n");        else if(n==2)            printf("1\n");        else        {            mem(arr.mat,0);            mem(tmp.mat,0);            arr.mat[0][0]=1,arr.mat[0][1]=1,arr.mat[1][0]=1,arr.mat[1][1]=0;            tmp.mat[0][0]=1,tmp.mat[0][1]=1,tmp.mat[1][0]=1,tmp.mat[1][1]=0;            Matrix p=pow_matrix(tmp,n-1);            p=mul(arr,p);            LL ans=(p.mat[1][0]+mod)%mod;            printf("%lld\n",ans);        }    }    return 0;}