nyoj 517 最小公倍数 【java睑板】

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:http://acm.nyist.net/JudgeOnline/problem.php?

pid=517

原文地址:https://www.cnblogs.com/mengfanrong/p/5023911.html