hdu2407

//公式为：((n-2)*(n-4)*(n-6)*(n-8)*(n-10)*........)/((n-1)*(n-3)*(n-5)*(n-7)*(n-9)*........) #include<iostream> #include<cstdio> #include<cmath> using namespace std; int main() { int n; while(scanf("%d",&n)!=EOF) { int i; double ans=1.0; for(i=2;i<=n-2;i+=2) ans*=(double)i/(double)(i+1); printf("%.5lf/n",ans); } return 0; }