A - 费马小定理

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

Input

输入一个数N(0 <= N <= 10^9)

Output

输出：计算结果

Sample Input

3

Sample Output

40

#include
using namespace std;
#define c 1000000007
long long f(long long a, long long b){
    int ans = 1;
    while(b){
        if(b & 1){
            ans = (ans * a)%c;
        }
        b>>=1;
        a = (a * a) % c;
    }
    return ans;
}
int main(){
    //long long sum = 0;
    int n;
    cin >> n;
    long long t = (f(3, n + 1) - 1);
   // cout << t << endl;
   // cout << f(2, c - 20000000) << endl;
    long long b1 = t * f(2, c - 2) % c;
    cout << b1 << endl;
    return 0;
}

 

一个数有逆元的充分必要条件是gcd(a,n)=1gcd(a,n)=1，此时逆元唯一存在 
(a b)mod p=((a mod p)×(b mod p) mod p