【POJ 3070】Fibonacci（矩阵快速幂）

【POJ 3070】Fibonacci（矩阵快速幂）

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:

.

Source

Stanford Local 2006

矩快基础题，训练计划里的居然一直没看。。。或者是对数论优先忽略了……

贴上来防止时间久了忘掉，还能过来看看

代码如下:

#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e4;
const double eps = 1e-8;

int ans[2][2];
int a[2][2];
int tmp[2][2];

void pow_m(int b)
{
	for(int i = 0; i < 2; ++i)
		for(int j = 0; j < 2; ++j)
		{
			ans[i][j] = !(i^j);
			a[i][j] = !(i&j);
		}

	while(b)
	{
		if(b&1)
		{
			memset(tmp,0,sizeof(tmp));
			for(int i = 0; i < 2; ++i)
				for(int j = 0; j < 2; ++j)
					for(int k = 0; k < 2; ++k)
						tmp[i][j] = (tmp[i][j] + ans[i][k]*a[k][j])%mod;
			memcpy(ans,tmp,sizeof(tmp));
		}

		b >>= 1;

		memset(tmp,0,sizeof(tmp));
		for(int i = 0; i < 2; ++i)
			for(int j = 0; j < 2; ++j)
				for(int k = 0; k < 2; ++k)
					tmp[i][j] = (tmp[i][j] + a[i][k]*a[k][j])%mod;
		memcpy(a,tmp,sizeof(tmp));
	}
}

int main()
{
	//fread();
	//fwrite();

	int n;
	while(~scanf("%d",&n) && n != -1)
	{
		pow_m(n);
		printf("%d\n",ans[0][1]);
	}

	return 0;
}