NYOJ 1103 区域赛系列一多边形划分

                                                                  区域赛系列一多边形划分

时间限制：1000 ms  |  内存限制：65535 KB

难度：2

描述

Give you a convex（凸边形）, diagonal n-3 disjoint divided into n-2 triangles(直线), for different number of methods, such as n=5, there are 5 kinds of partition method, as shown in Figure

 

输入

The first line of the input is a n (1<=n<=1000), expressed n data set.
The next n lines each behavior an integer m (3<=m<=18), namely the convex edges.

输出

For each give m,, output how many classification methods.
example output: Case #a : b

样例输入

3345

样例输出

Case #1 : 1Case #2 : 2Case #3 : 5

提示

Catalan number

*卡塔兰数是组合数学中一个常在各种计数问题中出现的数列。以比利时的数学家欧仁·查理·卡塔兰 (1814–1894)命名。 
卡塔兰数的一般项公式为 C_n = \frac{1}{n+1}{2n \choose n} = \frac{(2n)!}{(n+1)!n!} 
卡特兰数前几项为 :1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012……* 
递归式：令h(0)=1,h(1)=1,卡特兰数满足递归式：h(n)=h(0)*h(n-1)+h(1)*h(n-2)+…+h(n-1)*h(0); (n>=2) 
另类递归式：h(n)=((4*n-2)/(n+1))*h(n-1); 

#include <iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>using namespace std;int main(){   int t,ans=1,n;   double s,a[20];   scanf("%d",&t);   while(t--)   {       a[0]=1;       a[1]=1;       scanf("%d",&n);       for(int i=1;i<=n-2;i++)       {           s=(4*i-2)/((i+1)*1.0);           a[i]=s*a[i-1];       }       printf("Case #%d : %.0lf\n",ans++,a[n-2]);   }   return 0;}