HDU 2256 Problem of Precision （矩陣快速冪），hdu2256

HDU 2256 Problem of Precision （矩陣快速冪），hdu2256
HDU 2256 Problem of Precision （矩陣快速冪）

ACM

題目地址：HDU 2256 Problem of Precision

題意： 
給出一個式子，求值。

分析： 
推起來最後那步會比較難想。 
具體過程見： 
 
表示共軛只聽說過複數的和圖的... 
這構題痕迹好明顯... 
跟基友開玩笑說：如果遇到這種題，推到Xn+Yn*sqrt(6)這步時，打表最多隻能打到10就爆int了，這是輸出正解和Xn，說不定眼神好能發現ans = Xn * 2 - 1呢。= =...

代碼：

/**  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 matrixconst int MOD = 1024;struct Mat{int n;ll v[SIZE][SIZE];// value of matrixMat(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;}