HDU 4002 find the maximum (number theory-Euler's function)

Find the maximum
Problem descriptioneuler's totient function, Phi (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, Phi (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 PHI (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/PHI (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.
Inputthere are t test cases (1 <= T <= 50000). For each test case, standard input contains a line with 2 ≤ n ≤ 10 ^ 100.
Outputfor each test case there shoshould be single line of output answering the question posed above.
Sample Input

210100

 
Sample output

630HintIf the maximum is achieved more than once, we might pick the smallest such n.   

 
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Question:

Given an N, ask you 1 ~ In N, calculate a number X to maximize the value of X/PHI (X.

Solution:

According to the Euler's function formula, Phi (x) = x (1-1/P1) (1-1/P2) (1-1/P3) (1-1/P4) ..... (1-1/PN)

Then: X/PHI (x) = p1/(p1-1) * P2/(p2-1) *... * pn/(pn-1)

We can see that the more items x/PHI (x), the greater the factor, and the smaller the X/PHI (x), the greater the number, the only difference is 2*3*5*7 ....

Considering that the numbers are large, you can use Java to write them. Is it because of the large Java numbers, you know ..

Solution code:

import java.util.*;import java.math.*;public class Main {public static void main(String[] args) {int maxn=2100;boolean []isPrime =new boolean[maxn];int tol=0;int []v=new int[210];for(int i=0;i<maxn;i++) isPrime[i]=true;  for(int i=2;i<maxn;i++){if(tol>56) break;        if(isPrime[i]) v[tol++]=i;        for(int j=0;j<tol && i*v[j]<maxn;j++){              isPrime[i*v[j]]=false;              if(i%v[j]==0) break;          }}//System.out.println(tol);//for(int i=0;i<tol;i++) System.out.println(v[i]);BigInteger []a=new BigInteger[200];a[0]=new BigInteger("1"); for(int i=1;i<56;i++){BigInteger tmp= BigInteger.valueOf(v[i-1]) ;//System.out.print(tmp+" ");a[i]=a[i-1].multiply(tmp);//System.out.println(a[i]);}Scanner scan=new Scanner(System.in);int t=scan.nextInt();while(t-- >0){String str=scan.next();BigInteger x=new BigInteger(str);BigInteger ans=new BigInteger("2");for(int i=1;i<56;i++){if(x.compareTo(a[i]) >=0){ans=a[i];}else break;}System.out.println(ans);}}}