HDU 2256 Problem of Precision (matrix power), hdu2256

HDU 2256 Problem of Precision (matrix power), hdu2256
HDU 2256 Problem of Precision (matrix fast power)

ACM

Address: HDU 2256 Problem of Precision

Question:
Returns a formula and calculates the value.

Analysis:
It is hard to think about the last step.
For details, see:
 
It indicates that the person who has heard of the complex number and the graph only has the concatenation...
The question mark is obvious...
Joke with kiyou: If you encounter such a problem, when you push it to Xn + Yn * sqrt (6), you can only hit 10 at most to blow the int, which is the output positive solution and Xn, maybe we can see that ans = Xn * 2-1. =...

Code:

/**  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;}