杭电4002 亚洲区预赛大连赛区

这道题先是用数学公式可以推出来一个式子，接下来就是大数，，，碰到好几次大数了，以前都是用c或c++写的，这道题用c或c++实在是太麻烦，于是一咬牙，一跺脚，硬着头皮看起了Java，上午看了一点，下午问了问xd一些基本语法，于是便开始写了，既借助网上代码，又不断的问xd，终于ac了，这叫一个不容易啊，，用Java写的第一道题啊，，，，接下来再练一下Java，把nyoj上的大数题用Java再做一遍。。。。。。题目：

Euler's Totient function, φ (n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n . For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
HG is the master of X Y. One day HG wants to teachers XY something about Euler's Totient function by a mathematic game. That is HG gives a positive integer N and XY tells his master the value of 2<=n<=N for which φ(n) is a maximum. Soon HG finds that this seems a little easy for XY who is a primer of Lupus, because XY gives the right answer very fast by a small program. So HG makes some changes. For this time XY will tells him the value of 2<=n<=N for which n/φ(n) is a maximum. This time XY meets some difficult because he has no enough knowledge to solve this problem. Now he needs your help.

Input

There are T test cases (1<=T<=50000). For each test case, standard input contains a line with 2 ≤ n ≤ 10^100.

Output

For each test case there should be single line of output answering the question posed above.

Sample Input

2 10 100

Sample Output

6 30

ac代码：

import java.util.*; import java.math.BigInteger; public class Main { /** * @param args */ public static boolean pri(int n){ for(int i=2;i<=(int)Math.sqrt(n);++i) { if(n%i==0) return false; } return true; } public static void main(String[] args) { // TODO Auto-generated method stub BigInteger aa[]=new BigInteger[110]; int j=0,tt; int bb[]=new int[110]; Scanner cin=new Scanner(System.in); for(int i=2;j<110;++i) { if(pri(i)) { bb[j++]=i;} } aa[1]=new BigInteger("6"); BigInteger b6=new BigInteger("6"); //BigInteger b2=new BigInteger("2"); for(int i=2;i<105;++i) { aa[i]=new BigInteger(""+bb[i]).multiply(aa[i-1]); } BigInteger n; tt=cin.nextInt(); while(tt-->0) { n=cin.nextBigInteger(); if(n.compareTo(b6)<0) { if(n.compareTo(BigInteger.ONE)==0) System.out.println("1"); else System.out.println("2"); } else if(n.compareTo(b6)==0) System.out.println("6"); else{ for(int i=1;i<102;++i) { if(n.compareTo(aa[i])<=0) { System.out.println(aa[i-1]); break; } } } } } }