【矩阵快速幂】 斐波那契数列求解。

Fibonacci

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 11123 Accepted: 7913

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:

.

题意：给你一个求解第N个斐波那契数的公式。 让你求出Fn % 10000。

Solution：

都要比赛了，居然矩阵快速幂写得如此的不熟。。

思路：构造单位矩阵，使用矩阵快速幂。

PS

写矩阵乘法的时侯。

Matrix matrix_mul(Matrix A,Matrix B){
Matrix C;
memset(C.m,0,sizeof(C.m));
for (int i=0;i<=1;i++)
for (int k=0;k<=1;k++){
int r=A.m[i][k];
for (int j=0;j<=1;j++)
C.m[i][j]=(C.m[i][j]+(r*B.m[k][j])%M)%M;
}
return C;
}

这样的写法比下面的写法会快一些~

matrix multi(matrix a, matrix b)
{
matrix tmp;
for(int i = 0; i < 2; ++i)
{
for(int j = 0; j < 2; ++j)
{
tmp.m[i][j] = 0;
for(int k = 0; k < 2; ++k)
tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
}
}
return tmp;
}

为什么呢。

大神说这是缓存读取数据比内存读取数据快得多的原因，

还不不懂。

www.jb51.net/aricle/36422.htm

这个地方有详细的解释~

接下来就是一个基本的矩阵快速幂，

注意事项都在程序里面写了，

注意多写写，增加熟练度。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct Matrix{
int m[2][2];
}ans,base;
int n;
void test(){
for (int i=0;i<=1;i++){
for (int j=0;j<=1;j++)
cout<<ans.m[i][j]<<' ';
cout<<endl;
}
}
/*
*This way to improve the efficiency
*/
const int M=1e4;

Matrix matrix_mul(Matrix A,Matrix B){
Matrix C;
memset(C.m,0,sizeof(C.m));
for (int i=0;i<=1;i++)
for (int k=0;k<=1;k++){
int r=A.m[i][k];
for (int j=0;j<=1;j++)
C.m[i][j]=(C.m[i][j]+(r*B.m[k][j])%M)%M;
}
return C;
}
/*
*The n_th power of matrix base.

*initialize base and initialize ans as identify matix.

*<pay attention here>
*/
int power_matrix_mod(int n)
{
base.m[1][1]=0;
base.m[0][0]=base.m[0][1]=base.m[1][0]=1;
ans.m[0][0]=ans.m[1][1]=1;
ans.m[0][1]=ans.m[1][0]=0;
while(n){
if(n&1) ans=matrix_mul(ans,base);
base=matrix_mul(base,base);
n>>=1;
}
return ans.m[0][1];
}
int main()
{
scanf("%d",&n);
printf("%d ",power_matrix_mod(n));
return 0;
}