Covering(矩阵快速幂）

原创于 2019-04-10 22:24:03 发布 · 377 阅读

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在无限数量的1x2和2x1地毯的帮助下，Bob试图完全覆盖4xn的操场，而不让地毯重叠。本博客探讨了如何计算在不重叠条件下完全覆盖操场的所有方案的数量，使用递推公式和矩阵快速幂算法。

Bob's school has a big playground, boys and girls always play games here after school.

To protect boys and girls from getting hurt when playing happily on the playground, rich boy Bob decided to cover the playground using his carpets.

Meanwhile, Bob is a mean boy, so he acquired that his carpets can not overlap one cell twice or more.

He has infinite carpets with sizes of 1×21×2 and 2×12×1, and the size of the playground is 4×n4×n.

Can you tell Bob the total number of schemes where the carpets can cover the playground completely without overlapping?

Input

There are no more than 5000 test cases.

Each test case only contains one positive integer n in a line.

1≤n≤10181≤n≤1018

Output

For each test cases, output the answer mod 1000000007 in a line.

Sample Input

1
2

Sample Output

1
5

递推公式为F[N]=F[N-1]+5F[n-2]+F[n-3]-F[n-4];

这个题有个坑点，就是在取模之后数值会变小，然后导致系数为-的给减的导致最后的值是-的

这种问题我们的解法是在最后的结果+MOD再取模

学到了

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
#define MOD 1000000007
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
struct mat
{
  ll a[5][5];
};
mat  Mul(mat a,mat b)
{
	mat ans;
	memset(ans.a,0,sizeof(ans.a));
	for(int t=1;t<=4;t++)
	{
		for(int j=1;j<=4;j++)
		{
			for(int k=1;k<=4;k++)
			{
				ans.a[t][j]=(ans.a[t][j]+a.a[t][k]*b.a[k][j])%MOD;
			}
		}
	}
	return ans;

}
mat ans;
ll quickpow(ll n)
{
   mat res;
   memset(res.a,0,sizeof(res.a));
   res.a[1][1]=1;
   res.a[1][2]=5;
   res.a[1][3]=1;
   res.a[1][4]=-1;
   res.a[2][1]=1;
   res.a[3][2]=1;
   res.a[4][3]=1;
   while(n)
   {
   	if(n&1)
   	{
   		ans=Mul(res,ans);
	}
	res=Mul(res,res);
	n>>=1;
   }
   return ans.a[1][1];
}
int main()
{
	ll n;
	while(cin>>n)
	{
		
		memset(ans.a,0,sizeof(ans.a));
	    ans.a[1][1]=36;
	    ans.a[2][1]=11;
	    ans.a[3][1]=5;
	    ans.a[4][1]=1;
	    if(n==1)
	    {
	    	printf("1\n");
		}
		else if(n==2)
	    {
	    	printf("5\n");
		}
		else if(n==3)
		{
			printf("11\n");
		}
		else if(n==4)
		{
			printf("36\n");
		}
		else 
		{
			ll s=quickpow(n-4)+MOD;
			printf("%lld\n",(s)%MOD);
		}
		
	}

   return 0;
}