CodeForces 185A Plant（公式推导+快速幂）

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 nyears.

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).

Example

<
4000
div class="title" style="background-color:rgb(255,255,255);border-bottom:1px solid rgb(136,136,136);font-family:arial;font-weight:bold;padding:.25em;">
Input

1

Output

3

Input

2

Output

10

Note

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

题解：

观察规律可以看出在下一个图形变换的时候，对于现在每个朝上的三角形，下一次会变换成3个朝上的三角形和1个向下的三角形，对每个朝下的三角形下次会变成3个朝下的三角形和一个向上的三角形，这样就可以得出公式，假设每一次朝上三角形个数为a[i],朝下为b[i]，可以列出a[i]=a[i-1]*3+b[i-1]，和b[i]=3*b[i-1]+a[i-1]，这两个公式化简可以得到a[i]的公式为2的n-1次方加上2的2*n-1次方，这个用快速幂就可以处理了

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#include<stdio.h>
using namespace std;
#define ll long long
ll mod=1e9+7;
ll quick(ll x,ll y)
{
ll ans=1;
while(y!=0)
{
if(y%2==1)
ans=(ans*x)%mod;
x=(x*x)%mod;
y/=2;
}
return ans;
}
int main()
{
ll n;
scanf("%lld",&n);
if(n==0)
{
printf("1\n");
return 0;
}
printf("%I64d\n", (quick(2,n-1)+quick(2,2*n-1))%mod);
return 0;
}