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HDU 2256 Problem of Precision (矩阵快速幂)

HDU 2256 Problem of Precision (矩阵快速幂)

ACM

题目地址:HDU 2256 Problem of Precision

题意: 
给出一个式子,求值。

分析: 
推起来最后那步会比较难想。 
具体过程见: 
pic 
表示共轭只听说过复数的和图的... 
这构题痕迹好明显... 
跟基友开玩笑说:如果遇到这种题,推到Xn+Yn*sqrt(6)这步时,打表最多只能打到10就爆int了,这是输出正解和Xn,说不定眼神好能发现ans = Xn * 2 - 1呢。= =...

代码

/*
*  Author:      illuz <iilluzen[at]gmail.com>
*  Blog:        http://blog.csdn.net/hcbbt
*  File:        2256.cpp
*  Create Date: 2014-08-03 14:27:15
*  Descripton:   
*/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
#define repf(i,a,b) for(int i=(a);i<=(b);i++)
typedef long long ll;

const int SIZE = 3;		// max size of the matrix
const int MOD = 1024;

struct Mat{
	int n;
	ll v[SIZE][SIZE];	// value of matrix

	Mat(int _n = SIZE) {
		n = _n;
		memset(v, 0, sizeof(v));
	}

	void init(ll _v) {
		repf (i, 0, n - 1)
			v[i][i] = _v;
	}

	void output() {
		repf (i, 0, n - 1) {
			repf (j, 0, n - 1)
				printf("%lld ", v[i][j]);
			puts("");
		}
		puts("");
	}
} a, b;

Mat operator * (Mat a, Mat b) {
	Mat c(a.n);
	repf (i, 0, a.n - 1) {
		repf (j, 0, a.n - 1) {
			c.v[i][j] = 0;
			repf (k, 0, a.n - 1) {
				c.v[i][j] += (a.v[i][k] * b.v[k][j]) % MOD;
				c.v[i][j] %= MOD;
			}
		}
	}
	return c;
}

Mat operator ^ (Mat a, ll k) {
	Mat c(a.n);
	c.init(1);
	while (k) {
		if (k&1) c = c * a;
		a = a * a;
		k >>= 1;
	}
	return c;
}

void solve(int n) {
	Mat a(2);
	a.v[0][0] = 5;
	a.v[0][1] = 12;
	a.v[1][0] = 2;
	a.v[1][1] = 5;

	a = a ^ (n - 1);
	int xn = 5 * a.v[0][0] + 2 * a.v[0][1];
	printf("%d\n", (xn * 2 - 1) % MOD);
}

int t, n;

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		solve(n);
	}
	return 0;
}