UVa 1646 (递推 JAVA大数) Edge Case

题意:

有n个点围成一圈,这n个点的匹配就是没有公共点的边集(这些边只能连接一圈中相邻的两点),求所有匹配的个数。

额,我不会分析。。=_=||

算了几个数,找找规律发现它满足斐波那契数列的递推关系,f(n) = f(n-1) + f(n-2)

自从会用了Java的BigInteger,就懒得写C的高精度了。

 1 import java.io.*;
 2 import java.util.*;
 3 import java.math.*;
 4 
 5 public class Main
 6 {
 7     static PrintWriter out = new PrintWriter(new  BufferedWriter(new OutputStreamWriter(System.out)));
 8 
 9     public static void main(String args[]) throws IOException
10     {
11         BigInteger [] a = new BigInteger[10002];
12         a[3] = new BigInteger("4");
13         a[4] = new BigInteger("7");
14         for(int i = 5; i <= 10000; ++i) a[i] = a[i-1].add(a[i-2]);
15         Scanner cin=new Scanner(System.in);
16         while(cin.hasNext())
17         {
18             int n = cin.nextInt();
19             System.out.println(a[n]);
20         }
21     }
22 }
代码君
原文地址:https://www.cnblogs.com/AOQNRMGYXLMV/p/4289341.html