HDU - 2842 Chinese Rings

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Description

Dumbear likes to play the Chinese Rings (Baguenaudier). It’s a game played with nine rings on a bar. The rules of this game are very simple: At first, the nine rings are all on the bar.
The first ring can be taken off or taken on with one step.
If the first k rings are all off and the (k + 1)th ring is on, then the (k + 2)th ring can be taken off or taken on with one step. (0 ≤ k ≤ 7)

Now consider a game with N (N ≤ 1,000,000,000) rings on a bar, Dumbear wants to make all the rings off the bar with least steps. But Dumbear is very dumb, so he wants you to help him. 

Input

Each line of the input file contains a number N indicates the number of the rings on the bar. The last line of the input file contains a number "0". 

Output

For each line, output an integer S indicates the least steps. For the integers may be very large, output S mod 200907. 

Sample Input

 140 

 

Sample Output

 110 

題意：給定n個環和規則：如果想取下第n個環那麼要保證前n-2都取下，第n-1還在

思路：設取下第n個環的最短時間是f[n]，那麼要想取下第n個環，首先要取下前n-2個即：f[n-2]以及最後一個，所以是

f[n-2]+1， 還有一個第n-1個，要取下它首先要保證第n-2在，所以需要f[n-2]（怎麼取下來的怎麼放上去），現在又要取第n-1個了， 綜上所述：f[n] = 2*f[n-2] + f[n-1] + 1, 然後就是構造矩陣了：

不難推出來： | f[n] |              | 1 2 1 |     | f[n-2]|

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

                       | 1 |                | 0 0 1 |      |    1   |

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;const int maxn = 10;const int mod = 200907;int cnt;struct Matrix {int v[maxn][maxn];Matrix() {}Matrix(int x) {init();for (int i = 0; i < maxn; i++) v[i][i] = x;}void init() {memset(v, 0, sizeof(v));}Matrix operator *(Matrix const &b) const {Matrix c;c.init();for (int i = 0; i < cnt; i++)for (int j = 0; j < cnt; j++)for (int k = 0; k < cnt; k++)c.v[i][j] = (c.v[i][j] + (ll)v[i][k]*b.v[k][j]) % mod;return c;}Matrix operator ^(int b) {Matrix a = *this, res(1);while (b) {if (b & 1)res = res * a;a = a * a;b >>= 1;}return res;}} a, b, tmp;int main() {int n;while (scanf("%d", &n) != EOF && n != 0) {if (n < 3) {printf("%d\n", n);continue;}a.init();cnt = 3;a.v[0][0] = a.v[0][2] = a.v[1][0] = a.v[2][2] = 1;a.v[0][1] = 2;tmp = a^(n-2);printf("%d\n", (tmp.v[0][0]*2 + tmp.v[0][1] + tmp.v[0][2]) % mod);}return 0;}