AcWing 889. 满足条件的01序列
原题链接
简单
作者:
芝居気
,
2021-12-17 14:56:11
,
所有人可见
,
阅读 158
#include <iostream>
using namespace std;
typedef long long ll;
const int N = 200010;
const int mod = 1e9 + 7;
int fact[N], infact[N];
int qpow(int a, int b, int p)
{
int res = 1;
while (b)
{
if (b & 1) res = (ll)res * a % p;
a = (ll)a * a % p;
b >>= 1;
}
return res;
}
int C(int a, int b)
{
return (ll)fact[a] * infact[b] % mod * infact[a - b] % mod;
}
int ksm(int a, int k) {
int res = 1;
while (k) {
if (k & 1) res = (ll)res * a % mod;
a = (ll)a * a % mod;
k >>= 1;
}
return res;
}
int main() {
fact[0] = infact[0] = 1;
for (int i = 1; i < N; i++){
fact[i] = (ll)fact[i - 1] * i % mod;
infact[i] = (ll)infact[i - 1] * qpow(i, mod - 2, mod) % mod;
}
int n;
cin >> n;
int res = (ll)C(2*n,n) * ksm(n + 1, mod - 2) % mod;
cout << res << endl;
return 0;
}