3的幂的和

Description

求：3^0 + 3^1 +...+ 3^(N) mod 1000000007。

 

Input

每行一个整数N(0 <= N <= 10^9)

 

Output

输出：计算结果

 

Sample Input

3

Sample Output

40

 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
inline int read()
{
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
    while('0' <= ch && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
    return x * f;
}
const int mod=1000000007;
ll mod_pow(ll n)
{
    ll k=3,ans=1;
    while (n)
    {
        if (n&1) ans=(ans*k)%mod;
        n/=2;
        k=(k*k)%mod;
    }
    return ans;
}
int main()
{
    ll n,m;
    while(cin>>n){
    m=(mod+1)/2;
    ll ans=((mod_pow(n+1)-1)*m)%mod;
    cout<<ans<<endl;
    }
    return 0;
}