Fibonacci【矩阵乘法】（POJ 3070）

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:

.

---

 矩乘讲解：

直接递推计算时间复杂度显然为O(n)，而本题 n<=1e9，直接递推显然超时

我们知道，要求出 Fib(n) ，我们只需要知道Fib(n-1) 和 Fib(n-2)，我们在递推时只需要保存最近的两个斐波那契数即可

　　设 F(n) 表示一个1*2 的矩阵，F(n) = { F(n) , F(n+1) }

　　我们希望根据 F(n-1) = { F(n-1) , F(n) } 计算出 F(n) 。我们设一个矩阵 A  {0,1},{1,1}，那么：F(n) =F(n-1)*A=F(1)*A^n-1

　　A的次方，不就可以用快速幂优化吗？硬是把O(n) 优化为O( 2³log(n) ) 

 code

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long 
using namespace std;
const int mod=10000;
int k;

void mul(ll f[2],ll ju[2][2]) 
{
    ll c[2];memset(c,0,sizeof(c));
    for(int i=0;i<2;++i) 
        for(int j=0;j<2;++j) 
            c[i]=(c[i]+f[j]*ju[j][i])%mod;
    memcpy(f,c,sizeof(c));
}
void mulself(ll ju[2][2])
{
    ll c[2][2];memset(c,0,sizeof(c));
    for(int i=0;i<2;++i) // i 行 
        for(int j=0;j<2;++j) //j 列 
            for(int k=0;k<2;++k) 
                c[i][j]=(c[i][j]+ju[i][k]*ju[k][j])%mod; 
    memcpy(ju,c,sizeof(c));
}

int main()
{
    while(scanf("%d",&k)!=EOF)
    {
        if(k==-1) break;
        ll f[2]={0,1};
        ll ju[2][2]={{0,1},{1,1}};
        while(k) {
            if(k&1) mul(f,ju);
            mulself(ju);
            k>>=1;
        }
        printf("%lld
",f[0]);    
    }
    return 0;
}