演算法——矩陣快速冪 求第N個斐波那契數
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
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:
.
題意:給你一個求解第N個斐波那契數的公式。 讓你求出Fn % 10000。
構造單位矩陣,然後就是矩陣快速冪了。
快速冪的思路就是:
設A為矩陣,求A的N次方,N很大,1000000左右吧。。。
先看小一點的,A的9次方
A^9
= A*A*A*A*A*A*A*A*A 【一個一個乘,要乘9次】
= A*(A*A)*(A*A)*(A*A)*(A*A)【保持格式的上下統一,所以加上這句】
= A*(A^2)^4 【A平方後,再四次方,還要乘上剩下的一個A,要乘6次】
= A*((A^2)^2)^2【A平方後,再平方,再平方,還要乘上剩下的一個A,要乘4次】
也算是一種二分思想的應用吧,1000000次冪,暴力要乘1000000次,快速冪就只要(log2底1000000的對數) 次,大約20次
//
// main.cpp
// 斐波那契進階
//
// Created by 徐家興 on 2017/10/9.
// Copyright © 2017年 徐家興. All rights reserved.
//
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 3
#define LL long long
#define MOD 10007
using namespace std;
struct Matrix
{
LL a[MAXN][MAXN];
int r, c;//行數 列數
};
Matrix ori, res;//初始矩陣 和 結果矩陣
void init()//初始化矩陣
{
memset(res.a, 0, sizeof(res.a));
res.r = 2; res.c = 2;
for(int i = 1; i <= res.r; i++)//構造單位矩陣
res.a[i][i] = 1;
ori.r = 2; ori.c = 2;
ori.a[1][1] = ori.a[1][2] = ori.a[2][1] = 1;
ori.a[2][2] = 0;
}
Matrix multi(Matrix x, Matrix y)
{
Matrix z;
memset(z.a, 0, sizeof(z.a));
z.r = x.r, z.c = y.c;//新矩陣行數等於x矩陣的行數 列數等於y矩陣的列數
for(int i = 1; i <= x.r; i++)//x矩陣的行數
{
for(int k = 1; k <= x.c; k++)//矩陣x的列數等於矩陣y的行數 即x.c = y.r
{
if(x.a[i][k] == 0) continue;//優化
for(int j = 1; j<= y.c; j++)//y矩陣的列數
z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;
}
}
return z;
}
void Matrix_mod(long long n)
{
while(n)//N次冪
{
if(n & 1)
res = multi(ori, res);
ori = multi(ori, ori);
n >>= 1;
}
printf("%lld\n", res.a[1][2] % MOD);
}
int main()
{
long long N;
while(~scanf("%lld", &N))
{
init();//初始化單位矩陣
Matrix_mod(N);//矩陣快速冪
}
return 0;
}