题解 | 小红的双生排列

#include <iostream>
using namespace std;

typedef long long ll;
const int Mod  = 1e9 + 7;
ll n;

int main() {
    cin>>n;
    ll a = n/2 + (n&1);
    ll b = n/2;
    ll ans = 1;
    for(int i = 1;i<=a;i++) ans = (ans*i) % Mod;
    for(int i = 1;i<=b;i++) ans = (ans*i) % Mod;
    if(n%2==0) ans = (ans<<1) % Mod;
    cout<<ans;

    return 0;
}
// 64 位输出请用 printf("%lld")