AcWing 889. 满足条件的01序列
原题链接
简单
作者:
Drifter
,
2021-01-19 21:47:35
,
所有人可见
,
阅读 349
进阶 Day 4
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MOD = 1e9 + 7;
int qmi(int a, int k, int p)
{
int res = 1;
while (k)
{
if (k & 1) res = (LL)res * a % MOD;
k >>= 1;
a = (LL)a * a % MOD;
}
return res;
} //qmi
int main(void)
{
int n;
scanf("%d", &n);
int a = 2 * n, b = n;
int res = 1;
for (int i = a; i > a - b; i--) res = (LL)res * i % MOD;
for (int i = 1; i <= b; i++) res = (LL)res * qmi(i, MOD - 2, MOD) % MOD;
res = (LL)res * qmi(n + 1, MOD - 2, MOD) % MOD;
printf("%d\n", res);
return 0;
}