hdu 4002 Find the maximum 欧拉函数

Find the maximum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

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

题意：给你一个n，找到n/phi(n)最大的一个数，如果多个输出最小的那个；

思路：

        ；

　　　可得x/phi(x)=1/((1-1/p1)(1-1/p2).......(1-1/pn));

　　　令p最多，且最小即是答案；

　　　java码；

import java.util.*;
import java.math.*;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int[] a={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107
                ,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,
                337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,
                457,461,463,467,479,487,491,499,503,509,521,523,541};
        BigInteger[] ans=new BigInteger[110];
        ans[0]=new BigInteger("1");
        for(int i=1;i<=100;i++)
        ans[i]=ans[i-1].multiply(new BigInteger(""+a[i-1]));
        int T=cin.nextInt();
        for(;T!=0;T--)
        {
            BigInteger n=cin.nextBigInteger();
            for(int i=0;i<=100;i++)
            {
                if(ans[i].compareTo(n)>0)
                {
                    System.out.println(ans[i-1]);
                    break;
                }
            }
        }
    }
}