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hdu 4002 Find the maximum

时间:2014-05-08 08:33:37      阅读:299      评论:0      收藏:0      [点我收藏+]

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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

 

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 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 }
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hdu 4002 Find the maximum,布布扣,bubuko.com

hdu 4002 Find the maximum

标签:des   style   blog   class   code   java   

原文地址:http://www.cnblogs.com/tom987690183/p/3714999.html

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