POJ—3070—Fibonacci—【矩阵快速幂】

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Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9396 Accepted: 6668

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

099999999991000000000-1

Sample Output

0346266875

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:

Source

Stanford Local 2006

个赛有做过。。。用最朴素的方法，霸王硬上弓，结果超时了。。。。。。

后来听说是矩阵快速幂，那是完全不晓得什么是矩阵快速幂，后来隔了很久才学了一下。。。今天再来做这道题，1A，0Ms，感动哭了。。。

学过矩阵快速幂后写了一个专题，自转，发个链接：

http://blog.csdn.net/u013795055/article/details/38599321

解题思路其实就是学一下什么是矩阵快速幂，感觉是模板题。。。

【大神说：怎么还在做这道水题。。。o(╯□╰)o。。。给大神跪了】

直接上代码，感觉写的冗长了。。。可以优化的。。。

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>using namespace std;#define LL __int64#define MOD 10000typedef struct MATRIX   //定义一个矩阵结构体{    int mat[2][2];}MATRIX;MATRIX mat_multiply(MATRIX a,MATRIX b)  //矩阵乘法{    MATRIX c;    memset(c.mat,0,sizeof(c.mat));    for(int k=0;k<2;k++)        for(int i=0;i<2;i++)            if(a.mat[i][k])     //剪枝                for(int j=0;j<2;j++)                    if(b.mat[k][j]) //再剪枝，可以看上面的连接，完全一样的。。。                    {                        c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];                        if(c.mat[i][j]>=MOD)                            c.mat[i][j]%=MOD;   //取模                    }    return c;}MATRIX mat_quick_index (MATRIX a,int N){    MATRIX E;    for(int i=0;i<2;i++)    //学到的单位矩阵的酷炫炸天的初始化方法        for(int j=0;j<2;j++)            E.mat[i][j]=(i==j);    while(N)            //快速幂的核心    {        if(N &1)        //与运算，判断奇偶很有用的            E=mat_multiply(E,a);        N>>=1;          //位运算，向右是除以2，向左是乘以2，这个快！        a=mat_multiply(a,a);    }    return E;}int main(){    MATRIX a,b;    int N;    while(scanf("%d",&N)!=EOF)    {        if(N==-1)            break;        a.mat[0][0]=a.mat[0][1]=a.mat[1][0]=1;        a.mat[1][1]=0;        b=mat_quick_index(a,N);        printf("%d\n",b.mat[0][1]);    }    return 0;}

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