2 seconds
256 megabytes
standard input
standard output
Chris the Rabbit has been interested in arrays ever since he was a child. At the moment he is researching arrays with the length of n, containing only integers from 1 to n. He is not good at math, that's why some simple things drive him crazy. For example, yesterday he grew keen on counting how many different beautiful arrays there are. Chris thinks that an array is beautiful if it meets one of the two conditions:
- each elements, starting from the second one, is no more than the preceding one
- each element, starting from the second one, is no less than the preceding one
Having got absolutely mad at himself and at math, Chris came to Stewie and Brian to ask them for help. However, they only laughed at him and said that the answer is too simple and not interesting. Help Chris the Rabbit to find the answer at last.
The single line contains an integer n which is the size of the array (1 ≤ n ≤ 105).
You must print the answer on a single line. As it can be rather long, you should print it modulo 1000000007.
2
4
3
17
/***
题意：有一个长度为n的数字串，其中的数字都是1~n的数字，问：有多少种串能够满足下列两种情况：
非降序列；
非升序列。
两者等价，可以求出一种，然后另外一种就求出来了。 高中就能做出来，现在反而不会了= =！
借用模型，求出组合数学的公式，用数论的方法进行求解，如此即可。难点是数论上的方法求解。 先说模型：
高为1 ，长为2n-1的矩形，由1*1的小方格排列而成，然后，我们选取其中n-1个小方格，放上空格标志，
在剩下没有标志的空格里，记录从第一个小方格到当前小方格中，有多少个空格标志，将数量记录在方格中，
如此，满足非降序列条件的序列便可以得出，只不过，数字的范围是从0 ~ n-1，与从1 ~ n是等价的。 这种序列可以得到C(2n-1,n)个，同样的，满足非升序列的种数也是C(2n-1,n)，
但是，两种序列中有重复的：序列为0 0 0 0 0 0 …… 0，~ ，n-1，n-1，n-1，n-1，n-1，n-1，……n-1,
共有n中，
因此，
最终答案是，2C(2n-1,n)-n C(2n-1,n)mod p的求解： ((2n-1)!/n!*n!)mod p
先求n!*n!关于p的逆元m，
=((2n-1)!mod p)*(m mod p) mod p 而求n!*n!的逆元，可以根据费马小定理：a^(p-1) mod p = 1 (mod p)，成立的充分必要条件是a与p互质，
因此，当p为质数时（充分条件），等式仍成立。 逆元m满足 (n!*n!*m)mod p = 1恒成立，
已知的 p=1000000007,是质数，因此，可以根据费马小定理得到(n!*n!)*(n!*n!)^(p-2) mod p = 1,
因此，m= (n!*n!)^(p-2).
***/ #include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <vector>
#include <bitset>
#include <cstdio>
#include <string>
#include <numeric>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional> using namespace std;
typedef long long ll;
typedef unsigned long long ull; int dx[4]= {-1,1,0,0};
int dy[4]= {0,0,-1,1}; //up down left right
bool inmap(int x,int y,int n,int m)
{
if(x<1||x>n||y<1||y>m)return false;
return true;
}
int hashmap(int x,int y,int m)
{
return (x-1)*m+y;
} #define eps 1e-8
#define inf 0x7fffffff
#define debug puts("BUG")
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define lowbit(x) (-x)&x #define rep(i,n) for(int i=0;i<n;i++)
#define ll long long
#define mod 1000000007 int n;
ll res,denom; ll power(ll x,ll y)
{
ll ans = 1;
while(y)
{
if(y&1)
ans = ans * x %mod;
x = x*x%mod;
y>>=1;
}
return ans;
} int main()
{
cin>>n;
res = 1;
denom=1;
for(int i=2*n; i>n; i--)//2n*……*n+1
res = res * i%mod;
for(int i=1; i<=n; i++)//n!
denom = denom*i%mod;
denom = power(denom,(ll)mod-2);//变除为乘,求逆元
cout<<(res*denom-n+mod)%mod<<endl;
return 0;
}