Find the maximum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1990    Accepted Submission(s): 837

Problem Description

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

Hint

If the maximum is achieved more than once, we might pick the smallest such n.

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

题解：先算出来前100个数；找规律；由于数太大，用java；

代码：

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    static int vis[] = new int[1010];
    static int p[] = new int[1010];
    static BigInteger a[] = new BigInteger[110];
    static void getp()
    {
        for(int i = 0; i < 1010; i++)
            vis[i] = 0;
        vis[1] = 1;
        for(int i = 2; i <= 1000; i++)
        {
            if(vis[i] == 0)
            for(int j = i * i; j <= 1000; j += i)
            {
                vis[j] = 1;
            }
        }
        int tp = 0;
        for(int i = 1; i <= 1000; i++)
        {
            if(vis[i] == 0)
                p[tp++] = i;
        }
    }
    public static void main(String[] args){
        Scanner cin = new Scanner(System.in);
        getp();
        a[0]=BigInteger.valueOf(1);
        for(int i=1;i<=100;i++)
        {
            a[i] = a[i-1].multiply(BigInteger.valueOf(p[i-1]));
        }
        int t = cin.nextInt();
        while(t-- > 0)
        {
            BigInteger x;
            x = cin.nextBigInteger();
//            for(int i = 0; i <= 10; i++){
//                System.out.println(a[i]);
//            }
            if(x.compareTo(BigInteger.valueOf(6)) < 0){
                System.out.println("2");
                continue;
            }
            for(int i=0;i<=100;i++)
            {
                if(a[i].equals(x)){
                    System.out.println(a[i]);
                    break;
                }
                else if(a[i].compareTo(x) > 0)
                {
                    System.out.println(a[i-1]);
                    break;
                }
            }
        }
    }
}