Codeforces Round #324 (Div. 2) B. Kolya and Tanya （快速幂）

B. Kolya and Tanya

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.

More formally, there are 3n gnomes sitting in a circle. Each gnome can have from1 to3 coins. Let's number the places in the order they occur in the circle
by numbers from0 to3n - 1, let the gnome sitting on thei-th place haveai
coins. If there is an integeri (0 ≤ i < n) such thatai + ai + n + ai + 2n ≠ 6,
then Tanya is satisfied.

Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo109 + 7.
Two ways, a and b, are considered distinct if there is indexi (0 ≤ i < 3n), such thatai ≠ bi
(that is, some gnome got different number of coins in these two ways).

Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.

Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo109 + 7.

Sample test(s)

Input

1

Output

20

Input

2

Output

680

Note
20 ways for n = 1 (gnome with index0 sits on the top of the triangle, gnome1 on the right vertex, gnome
2 on the left vertex):


题意：一个环，上面有3n个点，一个点的值可以是1,2,3，问不存在a[i]+a[i+1]+a[i+2]=6这种情况的方案数

思路：列举可知道，存在那种情况的三个点为：123,132,213,222,231,312,321，共有7个，所以说

结果为3^(3*n)-7^n，快速幂一下就好了。。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#define MAXN 1010000
#define LL long long
#define ll __int64
#include<iostream>
#include<algorithm>
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
ll powmod(ll a,LL b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
ll a=powmod(3,3*n,1000000007);
ll b=powmod(7,n,1000000007);
//printf("%I64d %I64d\n",a,b);
printf("%I64d\n",(a-b+1000000007)%1000000007);
}
return 0;
}