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题目：
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Game of Connections |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 673 Accepted Submission(s): 443 |
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Problem DescriptionThis 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? |
| InputEach 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. |
| OutputFor each n, print in a single line the number of ways to connect the 2n numbers into pairs. |
Sample Input
2
3
-1 |
Sample Output
2
5 |
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| SourceAsia 2004, Shanghai (Mainland China), Preliminary |
| RecommendEddy |
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题目分析：
这一道题，读完题以后，抽象以下，其模型可以归结为“凸多边形的三角形划分。其中线段两两不相交”。而这一模型又可以使用卡特兰数来解决。
需要注意的是：catalans[20]就已经达到了6564120420。这已经超过了int的范围。catalans[99]的值为227508830794229349661819540395688853956041682601541047340。这已经超过了证书所能表示的范围，所以使用大数来处理
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代码如下：
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import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
BigInteger catalans[] = new BigInteger[101];
catalans[1] = new BigInteger(1);
BigInteger four = new BigInteger(4);
BigInteger two = new BigInteger(2);
BigInteger one = new BigInteger(1);
int i;
for(i = 2 ; i <= 100 ; ++i){//注意catalan[20]已经是6564120420了.所以需要用到大数
//根绝catalans数的递推公式来求得每一项的catalan数
catalans[i] = catalans[i-1].multiply(four.multiply(BigInteger.valueOf(i)).subtract(two)).divide(BigInteger.valueOf(i+1));
}
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()){
int n = scanner.nextInt();
if(n == -1){
return ;
}
System.out.println(catalans[n]);
}
}
}
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