hdu 4002 Find the maximum

Find the maximum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1561    Accepted Submission(s): 680

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

 

import java.io.*;
import java.awt.*;
import java.math.BigInteger;
import java.util.Scanner;

public class Main {

    static int prime[] = new int [1002];
    static int len = 0;
    static BigInteger dp[] = new BigInteger[1002]; 
    public static void main(String[] args) {
        fun();
        int T=0;
        Scanner cin  =  new Scanner(System.in);
        T=cin.nextInt();
        while(T>0)
        {
            BigInteger n = cin.nextBigInteger();
            int x=0;
            for(int i=1;i<=len;i++)
            {
                if(n.compareTo(dp[i])<0)
                {
                    x=i-1;
                    break;
                }
            }
            System.out.println(dp[x]);
            T--;    
        }
    }
    static void fun(){
        boolean s[] = new boolean[1007];
        for(int i=0;i<s.length;i++){
            s[i]=false;
        }
        for(int i=0;i<dp.length;i++){
            dp[i]=BigInteger.ZERO;
        }
        for(int i=2;i<1007;i++)
        {
            if(s[i]==true) continue;
            prime[++len]=i;
            for(int j=i*2;j<1007;j=j+i)
                s[j]=true;
        }
        dp[0] = BigInteger.ONE;
        for(int i=1;i<=len;i++)
        {
            dp[i] = dp[i-1].multiply(BigInteger.valueOf(prime[i]));
        }
    }
}