B. Kolya and Tanya

Time Limit: 1 Sec ?

Memory Limit: 256 MB

題目連接

http://codeforces.com/contest/584/problem/B

Description

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 from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n?-?1, let the gnome sitting on the i-th place have a_i coins. If there is an integer i (0?≤?i?<?n) such that a_i?+?a_i?+?n?+?a_i?+?2n?≠?6, then Tanya is satisfied.

Count the number of ways to choose a_i so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 10⁹?+?7. Two ways, a and b, are considered distinct if there is index i (0?≤?i?<?3n), such that a_i?≠?b_i (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 modulo 10⁹?+?7.

Sample Input

1

Sample Output

20

HINT

?

題意

給你一個(gè)環(huán)，環(huán)上有3n個(gè)點(diǎn)，每個(gè)點(diǎn)的權(quán)值可以是1-3，然后問你滿足a[i]+a[i+1]+a[i+2]！=6的方案有多少種

題解：

反面，a[i]+a[i+1]+a[i+2]=6的情況這三個(gè)數(shù)的取值一共有7種

那么答案就是 3^(3n) - 7^n就好了

代碼：

#include<stdio.h> #include<iostream> #include<math.h>using namespace std;#define mod 1000000007 long long quickpow(long long m,long long n,long long k) {long long b = 1;while (n > 0){if (n & 1)b = (b*m)%k;n = n >> 1 ;m = (m*m)%k;}return b; }int main() {long long n;cin>>n;cout<<(quickpow(3,3*n,mod) - quickpow(7,n,mod) + mod) % mod <<endl; }

?

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