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

 

 1 import java.io.*;

 2 import java.awt.*;

 3 import java.math.BigInteger;

 4 import java.util.Scanner;

 5 

 6 public class Main {

 7 

 8     static int prime[] = new int [1002];

 9     static int len = 0;

10     static BigInteger dp[] = new BigInteger[1002]; 

11     public static void main(String[] args) {

12         fun();

13         int T=0;

14         Scanner cin  =  new Scanner(System.in);

15         T=cin.nextInt();

16         while(T>0)

17         {

18             BigInteger n = cin.nextBigInteger();

19             int x=0;

20             for(int i=1;i<=len;i++)

21             {

22                 if(n.compareTo(dp[i])<0)

23                 {

24                     x=i-1;

25                     break;

26                 }

27             }

28             System.out.println(dp[x]);

29             T--;    

30         }

31     }

32     static void fun(){

33         boolean s[] = new boolean[1007];

34         for(int i=0;i<s.length;i++){

35             s[i]=false;

36         }

37         for(int i=0;i<dp.length;i++){

38             dp[i]=BigInteger.ZERO;

39         }

40         for(int i=2;i<1007;i++)

41         {

42             if(s[i]==true) continue;

43             prime[++len]=i;

44             for(int j=i*2;j<1007;j=j+i)

45                 s[j]=true;

46         }

47         dp[0] = BigInteger.ONE;

48         for(int i=1;i<=len;i++)

49         {

50             dp[i] = dp[i-1].multiply(BigInteger.valueOf(prime[i]));

51         }

52     }

53 }