NYOJ 164 Game of Connections



Game of Connections

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

This is a small but ancient game. You are supposed to write down the numbers 1, 2, 3, . . . , 2n - 1, 2n consecutively in clockwise order on the ground to form a circle, and then, to draw some straight line segments to connect them into number pairs. Every number must be connected to exactly one another. 
And, no two segments are allowed to intersect. 
It's still a simple game, isn't it? But after you've written down the 2n numbers, can you tell me in how many different ways can you connect the numbers into pairs? Life is harder, right?

输入

Each line of the input file will be a single positive number n, except the last line, which is a number -1. 
You may assume that 1 <= n <= 100.

输出

For each n, print in a single line the number of ways to connect the 2n numbers into pairs

样例输入

23-1

样例输出

25

来源

POJ

卡特兰数（列）：

令h(0)=1,h(1)=1，catalan数满足递推式：h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2)
例如：h(2)=h(0)*h(1)+h(1)*h(0)=1*1+1*1=2               h(3)=h(0)*h(2)+h(1)*h(1)+h(2)*h(0)=1*2+1*1+2*1=5
另类递推式：        h(n)=h(n-1)*(4*n-2)/(n+1);
递推关系的解为： h(n)=C(2n,n)/(n+1) (n=0,1,2,...)
递推关系的另类解为：    h(n)=c(2n,n)-c(2n,n-1)(n=0,1,2,...)

一些方面的应用：

1. 括号化:矩阵连乘:P=a1×a2×a3×……×an，依据乘法结合律，不改变其顺序，只用括号表示成对的乘积，试问有几种括号化的方案？(h(n-1)种)
2.一个栈(无穷大)的进栈序列为1，2，3，…，n，有多少个不同的出栈序列?
3.给定n个点求能组成的二叉树所有总数。
4. 凸多边形三角划分（任意两顶点之间的连线必能相交），求有多少中分割的方法（类似：在圆上选择2n个点,将这些点成对连接起来使得所得到的n条线段不相交的方法数）
5. n层阶梯切割为n个矩形的切割方法总数

代码：

#include<stdio.h>#include<iostream>using namespace std;int a[105][100];void catalan() //求卡特兰数,a[i][j]存储的是第i个逆序（高位在后）的卡特兰数（从0开始），且未对高位0进行处理 {    int i, j, len, carry, temp;    a[1][0] = 1;    len = 1;    for(i = 2; i <= 100; i++)    {        for(j = 0; j < len; j++) //乘法        a[i][j] = a[i-1][j]*(4*(i-1)+2);        carry = 0;        for(j = 0; j < len; j++) //处理相乘结果        {            temp = a[i][j] + carry;            a[i][j] = temp % 10;            carry = temp / 10;        }        while(carry) //进位处理        {            a[i][len++] = carry % 10;            carry /= 10;        }        carry = 0;        for(j = len-1; j >= 0; j--) //除法        {            temp = carry*10 + a[i][j];            a[i][j] = temp/(i+1);            carry = temp%(i+1);        }    }}int main(){int n;catalan() ;while(scanf("%d",&n) ,n != -1){int flag = 0 ;for(int i = 99;i >= 0;i--){if(a[n][i] != 0)flag = 1;if(flag)printf("%d",a[n][i]);}printf("\n");}return 0;}