HDU - 2842 Chinese Rings

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 1
4
0 

Sample Output

 1
10 

题意：给定n个环和规则：如果想取下第n个环那么要保证前n-2都取下，第n-1还在

思路：设取下第n个环的最短时间是f[n]，那么要想取下第n个环，首先要取下前n-2个即：f[n-2]以及最后一个，所以是

f[n-2]+1， 还有一个第n-1个，要取下它首先要保证第n-2在，所以需要f[n-2]（怎么取下来的怎么放上去），现在又要取第n-1个了， 综上所述：f[n] = 2*f[n-2] + f[n-1] + 1, 然后就是构造矩阵了：

不难推出来： | 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;
}

HDU - 2842 Chinese Rings,布布扣,bubuko.com

HDU - 2842 Chinese Rings

标签：des   style   blog   os   io   for   2014   ar   

原文地址：http://blog.csdn.net/u011345136/article/details/38305159