HDU 2842 Chinese Rings (with constant matrix + rapid matrix power), hdu2842

HDU 2842 Chinese Rings (with constant matrix + rapid matrix power), hdu2842
HDU 2842 Chinese Rings (with constant matrix + rapid matrix power)

ACM

Address: HDU 2842 Chinese Rings

Question:
A Chinese ring, the k ring to be unlocked first needs to be unlocked before (K-2) a ring, and leave the (k-1) ring. It takes at least a few steps to unlock n rings.

Analysis:
If f (n) is set, the n ring is unlocked.
1. Due to the game rules, the n-Ring cannot be completely unlocked at once, or n-1 n-Ring cannot be removed.
2. First remove the n: first complete f (n-2), and then remove the n ring.
3. Then put back before (n-2), in fact this is also f (n-2), because it is a inverse process.
4. Finally, f (n-1) is completed.

What about f (n-1? It's its own business...
So f (n) = f (n-2) + 1 + f (n-2) + f (n-1 ).
The first time I encountered a formula sub containing constants, the constants could have been used.

 
 

  
  
| f[n] | | 1 2 1 | |f[n-2]|
  
  
|f[n-1]| = | 1 0 0 | * |f[n-1]|
  
  
| 1 | | 0 0 1 | | 1 |
 
 

Code:

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        2842.cpp*  Create Date: 2014-08-03 22:22:57*  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 N = 31;const int SIZE = 3;// max size of the matrixconst int MOD = 200907;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) {memset(v, 0, sizeof(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, c;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;}ll solve(int n) {if (n <= 2) {return n;}// inita.v[0][1] = 2;a.v[0][0] = a.v[0][2] = a.v[1][0] = a.v[2][2] = 1;b.v[0][0] = 2;b.v[1][0] = b.v[2][0] = 1;c = a ^ (n - 2);c = c * b;return c.v[0][0];}int main() {int n;while (~scanf("%d", &n) && n) {cout << solve(n) << endl;}return 0;}