我写了一个gcd TL该。然后调用math内gcd，AC该、、、

思维：它是采取n前面的最小公倍数和n求 1~n的最小公倍数

代码：

import java.util.Scanner;

import java.math.*;

public class Main{

	public static void main(String[] args){

		Scanner cin = new Scanner(System.in);

		BigInteger[] s = new BigInteger[102];

		s[1] = new BigInteger("1");

		s[2] = new BigInteger("2");

		int i;

		for(i = 3; i < 102; i ++){

			s[i] = new BigInteger(((Integer)i).toString());

			BigInteger temp = s[i-1].gcd(s[i]);

			s[i] = s[i].multiply(s[i-1]).divide(temp);

			//System.out.println(s[i]);

		}

		int n;

		while(cin.hasNext()){

			n = cin.nextInt();

			System.out.println(s[n]);

		}

	}

}  

主题链接：

pid=517">http://acm.nyist.net/JudgeOnline/problem.php?

pid=517