POJ 3070 Fibonacci 【矩阵快速幂】

题目连接 ： http://poj.org/problem?id=3070

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Fibonacci

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 12738 Accepted: 9065

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:

.

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题目大意 ： 这个的大意应该不用说了吧

题解 ： 这个题目大意里已经说得很清楚了吧 。。

就是裸的矩阵快速幂 没什么难度 主要是用来熟悉下 矩阵快速么的代码实现 并顺手撸了一发模板

const int M = 2;
const int MOD = 1e4;

struct Matrix
{
LL m[M][M];
};

Matrix operator * (Matrix a,Matrix b)
{
Matrix c;
for(int i=0; i<M; i++) //初始化矩阵
for(int j=0; j<M; j++)
c.m[i][j]= 0;

for(int k=0; k<M; k++)
for(int i=0; i<M; i++) //实现矩阵乘法
{
if(a.m[i][k] <= 0)  continue;
for(int j=0; j<M; j++)
{
if(b.m[k][j] <= 0)    continue;
c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j]+MOD)%MOD;
}
}
return c;
}

Matrix operator ^ (Matrix a,LL b)
{
Matrix c;
for(int i=0; i<M; i++) //初始化单位矩阵
for(int j=0; j<M; j++)
c.m[i][j]= ( i == j );

while(b)
{
if(b&1) c= c * a ;
b >>= 1;
a = a * a ;
}

return c;
}

直接附本题代码吧

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#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <algorithm>
#include <math.h>

using namespace std;

#define LL long long int
#define _LL __int64

/******************************/

/*****************************/

const int M = 2;
const int MOD = 1e4;

struct Matrix
{
LL m[M][M];
};

Matrix operator * (Matrix a,Matrix b)
{
Matrix c;
for(int i=0; i<M; i++) //初始化矩阵
for(int j=0; j<M; j++)
c.m[i][j]= 0;

for(int k=0; k<M; k++)
for(int i=0; i<M; i++) //实现矩阵乘法
{
if(a.m[i][k] <= 0)  continue;
for(int j=0; j<M; j++)
{
if(b.m[k][j] <= 0)    continue;
c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j]+MOD)%MOD;
}
}
return c;
}

Matrix operator ^ (Matrix a,LL b)
{
Matrix c;
for(int i=0; i<M; i++) //初始化单位矩阵
for(int j=0; j<M; j++)
c.m[i][j]= ( i == j );

while(b)
{
if(b&1) c= c * a ;
b >>= 1;
a = a * a ;
}

return c;
}

void init(Matrix &a)
{
a.m[0][0]=1;
a.m[0][1]=1;
a.m[1][0]=1;
a.m[1][1]=0;
}
int main()
{
ios::sync_with_stdio(false);

Matrix f;
int n;
while(~scanf("%d",&n)&&n!=-1)
{
init(f);
f=f^n;

printf("%lld/n",f.m[0][1]);
}
return 0;
}