ACM-ICPC 2018 焦作赛区网络预赛 Poor God Water（递推+构造矩阵）

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 3 continuous hours when he eats only one kind of food, he will be unhappy. Besides, if there are 3 continuous hours when he eats all kinds of those, with chocolate at the middle hour, it will be dangerous. Moreover, if there are 3 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 1000000007.

Input

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

Each of the next T lines contains an integer NNthat shows the number of hours. (1≤N≤1010)

Output

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

样例输入复制

3
3
4
15

样例输出复制

20
46
435170

题目来源

ACM-ICPC 2018 焦作赛区网络预赛

题意：有肉，鱼，巧克力三种食物，有几种禁忌，对于连续的三个食物，1.这三个食物不能都相同；2.若三种食物都有的情况，巧克力不能在中间；3.如果两边是巧克力，中间不能是肉或鱼，求方案数

解析：起先递推枚举最后一位，1,2,3（2是巧克力）但是发现递推太多。就枚举后两位，11,12,13,21，.......

f(i,j,k):表示第i小时，最后倒数第二位是j，最后一位是k。

则f(i,1,1)=f(i-1,2,1)+f(i-1,3,1)

。。。（剩下8个递推）

#include<bits/stdc++.h>
using namespace std;
 
#define e exp(1)
#define pi acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}

const int N=9;
ll n;
struct mat
{
    ll a[N][N];
};
mat mat_mul(mat x,mat y)  
{
    mat res;  
    mem(res.a,0);  
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++) 
        for(int k=0;k<N;k++)
        res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j])%mod;
    return res;
}
void mat_pow(ll n)
{
    mat c,res;
    mem(c.a,0);
    
    c.a[0][3]=c.a[0][6]=1;
    c.a[1][0]=c.a[1][6]=1;
    c.a[2][0]=c.a[2][3]=c.a[2][6]=1;
    c.a[3][1]=c.a[3][4]=1;
    c.a[4][1]=c.a[4][7]=1;
    c.a[5][4]=c.a[5][7]=1;
    c.a[6][2]=c.a[6][5]=c.a[6][8]=1;
    c.a[7][2]=c.a[7][8]=1;
    c.a[8][2]=c.a[8][5]=1;
    
    mem(res.a,0);
    for(int i=0;i<N;i++) res.a[i][i]=1;  
    while(n)
    {
        if(n&1) res=mat_mul(res,c);  
        c=mat_mul(c,c);  
        n>>=1;
    }
    
    //res.a
    ll ans=0;
    for(int i=0; i<N; i++)
    {
    	for(int j=0; j<N; j++)
    	{
    		ans=(ans+res.a[i][j])%mod;
		}
	}
	printf("%lld\n",ans);
}
int main()
{
	int T;scanf("%d",&T);
	while(T--)
	{
		scanf("%lld",&n);
		if(n==1)puts("3");
		else if(n==2)puts("9");
		else
		{
			mat_pow(n-2);
		}
	}
	return 0;
}