POJ 3070——Fibonacci【矩阵快速幂的构造】

题目传送门

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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.

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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.

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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).

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Sample Input

0
9
999999999
1000000000
-1

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Sample Output

0
34
626
6875

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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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Source

Stanford Local 2006

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题意

• 利用矩阵快速幂求Fibonacci数列第 n 项

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题解

• 很简单的矩阵快速幂，可以按照题目给的矩阵构造，也可以像我这样构造，如下图
  

AC-Code

#include <bits/stdc++.h>
//#pragma GCC optimize("O3")
//#pragma G++ optimize("O3")
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define ll long long
#define RI register int 
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

const int maxn = 3;
const int mod = 1e4;
#define mod(x)	((x)%mod)

struct mat {
	int m[maxn][maxn];
	mat() {
		memset(m, 0, sizeof m);
	}
}unit;

mat operator * (mat a, mat b) {
	mat ret;
	ll x;
	for (ll i = 0; i < maxn; ++i)
		for (ll j = 0; j < maxn; ++j) {
			x = 0;
			for (ll k = 0; k < maxn; ++k)
				x += mod((ll)a.m[i][k] * b.m[k][j]);
			ret.m[i][j] = mod(x);
		}
	return ret;
}

void init_unit() {
	for (int i = 0; i < maxn; ++i)
		unit.m[i][i] = 1;
}

mat pow_mat(mat a, ll n) {
	mat ret = unit;
	while (n) {
		if (n & 1)	ret = ret * a;
		a = a * a;
		n >>= 1;
	}
	return ret;
}

int main() {
	ios;
	int n;
	init_unit();
	while (cin >> n && n != -1) {
		if (n == 0)		puts("0");
		else if (n == 1)	puts("1");
		else if (n == 2)		puts("1");
		else {
			mat a, b;
			b.m[0][0] = 1, b.m[0][1] = 1, b.m[0][2] = 0;
			b.m[1][0] = 1, b.m[1][1] = 0, b.m[1][2] = 1;
			b.m[2][0] = 0, b.m[2][1] = 0, b.m[2][2] = 0;

			a.m[0][0] = 1, a.m[0][1] = 1, a.m[0][2] = 0;
			b = pow_mat(b, n - 2);
			a = a * b;
			cout << mod(a.m[0][0]) << endl;
		}
	}
	return 0;
}