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

;
};
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;
}