ACM-ICPC 2018 焦作赛区网络预赛——Poor God Water(矩阵快速幂)

最新推荐文章于 2019-08-30 13:29:16 发布

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最新推荐文章于 2019-08-30 13:29:16 发布

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分类专栏：快速幂运算

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解决一道复杂的饮食安排问题，涉及连续小时的肉、鱼和巧克力食用，避免特定组合导致的不快或危险，通过矩阵快速幂算法高效计算安全饮食方案数量。

God Water likes to eat meat, fish and chocolate very much, but unfortunately, the doctor tells him that some sequence of eating will make them poisonous.

Every hour, God Water will eat one kind of food among meat, fish and chocolate. If there are 33 continuous hours when he eats only one kind of food, he will be unhappy. Besides, if there are 33 continuous hours when he eats all kinds of those, with chocolate at the middle hour, it will be dangerous. Moreover, if there are 33 continuous hours when he eats meat or fish at the middle hour, with chocolate at other two hours, it will also be dangerous.

Now, you are the doctor. Can you find out how many different kinds of diet that can make God Water happy and safe during NN hours? Two kinds of diet are considered the same if they share the same kind of food at the same hour. The answer may be very large, so you only need to give out the answer module 10000000071000000007.

Input

The fist line puts an integer TT that shows the number of test cases. (T \le 1000T≤1000)

Each of the next TT lines contains an integer NN that shows the number of hours. (1 \le N \le 10^{10}1≤N≤1010)

Output

For each test case, output a single line containing the answer.

样例输入复制

样例输出复制

20
46
435170

题目来源

ACM-ICPC 2018 焦作赛区网络预赛

题目链接:

传送门

解题思路：

当有3h我们可以很容易列举出一共20种可行方案（除去中毒的7种），当需要在3h基础上增加1h时，当前方案数 = 当前能到达结尾两种食物状态的数量 * (能到达下一个可行食物的方案总数) ，由于两种食物两两搭配只有9种情况，假设连续三种食物 M F C 那么就可以用M F 当做二维矩阵的行， F C当做列，这样就能简单地使用9*9矩阵表示所有初始状态，可行的方案可以初始为1

#include<bits/stdc++.h>
using namespace std;
typedef long long  ll;
const ll mod = 1000000007;
struct matrix
{
	ll a[10][10] = {{0}}; 
	matrix operator * (const matrix &b) const
	{
		matrix ret;
		for(int i = 1;i<=9 ; i++)
		{
			for(int j = 1; j<=9; j++)
			{
				 ret.a[i][j] = 0;
				 for(int k = 1;k<=9;k++)
				 {
				 	ret.a[i][j] += (a[i][k] * b.a[k][j])%mod; 
				 	ret.a[i][j]%=mod;
				 }
			}
		}
		return ret;
	}
};
matrix quick_pow(matrix t, ll n)
{

	n -= 3;
	matrix ret = t;
	matrix temp = t; 
	while(n)
	{
		if(n&1) ret = ret * temp; 
		n>>= 1;
		temp = temp * temp; 
	}
	return ret;
}
int main(void)
{
	int T;
	cin>>T;
	while(T--)
	{
	 	ll n;cin>>n;
	 	if(n<3) { cout<<pow(3,n)<<endl; continue;} 
	 	matrix m;	 	
	 	m.a[1][3] = m.a[1][2] = 1;
	 	m.a[2][4] = m.a[2][5] = m.a[2][6] = 1;
	 	m.a[3][7] = m.a[3][9] = 1;
	 	m.a[4][1] = m.a[4][2] = m.a[4][3] = 1;
	 	m.a[5][4] = m.a[5][6] = 1;
	 	m.a[6][9] = m.a[6][7] = 1;
	 	m.a[7][1] = m.a[7][2] = 1;
	 	m.a[8][4] = m.a[8][5] = 1;
	 	m.a[9][7] = m.a[9][8] = 1;    
        m = quick_pow(m,n); 
	  	ll ans = 0;         
		for (int i = 1; i <= 9; i++) 
		{                
			for (int j = 1; j <= 9; j++) 
			{
				ans += m.a[i][j] ;
				ans %= mod;
			}
		}
		cout<<ans<<endl;
	}
}