Codeforces 584 B. Kolya and Tanya (Codeforces Round #324 (Div. 2))


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 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 ai coins. If there is an integer i (0 ≤ i < n) such that ai + 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 modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ 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 modulo 109 + 7.

Sample test(s)
input
1
output
20
input
2
output
680
Note

20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex):


題目大意:

給一個n,表示一張桌子坐了3*n個人

你要給每一個人發錢,可以發1,2,3塊,第i個人的錢是ai  (i從0開始)

有一個要求:. ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.

也就是只要有 一組that ai + ai + n + ai + 2n ≠ 6,  .就可以了

解題思路:

對於n=1的時候,我們可以發現這個三角形有 20種情況符合條件,7種不符合條件;

我們可以這么想,反正只需要至少有一種情況滿足就行,所以我們就從它的對立事件想,

就是全部的數量減去它一個也不滿足的數量也就是

ret == 27 ^ n --  7 ^ n

只要推出這個來,只需要進行一下快速冪就ok了;

上代碼:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;

#define MM(a) memset(a,0,sizeof(a))
#define MM1(a) memset(a, false, sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e3+5;
const int mod = 1e9 + 7;
const double eps = 1e-7;
const double pi = 3.1415926;

LL quick_mod(LL a, LL b)
{
    LL ans = 1;
    while(b)
    {
        if(b&1)
            ans = (ans*a)%mod;
        b>>=1;
        a = (a*a)%mod;
    }
    return ans;
}
int main()
{
    int n;
    cin>>n;
    LL ans = quick_mod(27, n);
    LL ret = quick_mod(7, n);
    ///cout<<ans<<" " <<ret<<endl;
    ret = (ans%mod - ret%mod +mod)%mod;
    cout<<ret<<endl;
}



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