SGU130-circle - c++编程基础

130. Circle

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

On a circle border there are 2k different points A1, A2, ..., A2k, located contiguously. These points connect k chords so that each of points A1, A2, ..., A2k is the end point of one chord. Chords divide the circle into parts. You have to find N - the number of different ways to connect the points so that the circle is broken into minimal possible amount of parts P.

Input

The first line contains the integer k (1 <= k <= 30).

Output

The first line should contain two numbers N and P delimited by space.

Sample Input

Sample Output

2 3

题意是说有一个圆、给你2k个点，让你连接这些点的弦最少能把这个圆划分为几块、并且求出其方法数。

#include
      
       
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         #include
        
          #include
         
           #include
          
            #include
           
             #include
            
              #include
             
               #include
              
                #include
               
                 #include
                 #include
                 
                   #include
                  
                    using namespace std; int main() { int i,j,n; long long D[31]; D[0]=1; D[1]=1; D[2]=2; //画一条弦，则圆被分成两部分，两部分可以各自看成点数比较少的圆， //用两部分分割方法数相乘。以一点为这条弦的一端，枚举另一端求和。 for(i=3; i<=30; i++) { D[i]=0; for(j=1; j<=i; j++) D[i]+=D[j-1]*D[i-j]; } while(cin>>n) { cout<