求逆元 费马小

hdu 657

Rng

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 244    Accepted Submission(s): 154

 

Problem Description

Avin is studying how to synthesize data. Given an integer n, he constructs an interval using the following method: he first generates a integer r between 1 and n (both inclusive) uniform-randomly, and then generates another integer l between 1 and r (both inclusive) uniform-randomly. The interval [l, r] is then constructed. Avin has constructed two intervals using the method above. He asks you what the probability that two intervals intersect is. You should print p* q(−1)(MOD 1, 000, 000, 007), while pq denoting the probability.

 

 

Input

Just one line contains the number n (1 ≤ n ≤ 1, 000, 000).

 

 

Output

Print the answer.

 

 

Sample Input

1

2

 

Sample Output

1

750000006

 

 

逆元主要用于求除法取模 即求（a/b)%p 的值

当p为素数时（例1e9+7）可以用费马小定理求逆元：　　对于质数p，任意整数a，均满足：a^p≡a（mod p）

设整数为a，模数为 p，a在 mod p条件下的逆元为 x。

首先，我们由费马小定理得

　　a^(p - 1) ≡ 1 mod p

变形得

　　a * a^(p-2) ≡ 1 mod p

对比于逆元的定义

　　a * x ≡ 1 mod p

显然 我们可以得到

　　x ≡ a^(p-2) mod p

这样，我们就由费马小定理推理出 a 在 mod p 条件下的逆元为 a^(p-2)。

#include
using namespace std;
const int mod=1e9+7;
long long n;
long long ksm(long long x,long long y)
{
	long long res=1;
	while(y)
	{
		if(y%2)
		{
			res=res*x%mod;
		}
		x=x*x%mod;
		y/=2;
	}
	return res;
}
int main()
{
	while(cin>>n)
	{
//找规律得出公式（n+1）/(2*n)%mod 使用逆元得 （n+1）* x %mod（x 为 2*n 的逆元）
		cout<<(n+1)*ksm(2*n,mod-2)%mod<