CodeForces - 185A Plant

Description

Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will
point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats.
The figure below illustrates this process.



Help the dwarfs find out how many triangle plants that point "upwards" will be in n years.

Input

The first line contains a single integer n(0 ≤ n ≤ 1018) — the number of full years when the plant
grew.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output

Print a single integer — the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007(109 + 7).

Sample Input

Input

1

Output

3

Input

2

Output

10

Hint

The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.

Source

Codeforces Round #118 (Div. 1)


思路：矩阵快速幂

第n年的upwards的数量我们可以看成Xn,downwards的数量可以看成Yn,

那么有

X0=1, Y0=0

X1=3*X0+Y0, Y1=X0+3*Y0

.......

于是就有

Xn=3*X(n-1)+Y(n-1)

Yn=X(n-1)+3*Y(n-1)

那么接下来看矩阵如何操作

利用矩阵的特性：

X Y 3 1 3*X+Y X+3*Y

* =

Y X 1 3 X+3*Y 3*X+Y

正好满足我们上面的递推公式，说明我们的矩阵构造成功了

AC代码：

#include<iostream>
#include<string.h>
using namespace std;

typedef long long LL;

const LL N = 2;
const LL M = 1e9+7;

struct mat
{
LL a[2][2];
};

mat mul(mat a,mat b)
{
mat c;
memset(c.a,0,sizeof(c.a));
for(int i=0;i<N;i++)
for(int j=0;j<N;j++)
for(int k=0;k<N;++k)
{
c.a[i][j]+=(a.a[i][k]*b.a[k][j])%M;
c.a[i][j]%=M;
}
return c;
}

mat quick(mat a,LL n)
{
mat b;
b.a[0][0]=b.a[1][1]=1;
b.a[0][1]=b.a[1][0]=0;
while(n>0)
{
if(n&1)b=mul(a,b);
a=mul(a,a);
n>>=1;
}
return b;
}

int main()
{
LL n;
while(cin>>n)
{
mat a;
a.a[0][0]=a.a[1][1]=3;
a.a[1][0]=a.a[0][1]=1;
a=quick(a,n);
cout<<a.a[0][0]<<endl;
}
return 0;
}