題目：Maximum Number Of Divisors

import java.io.BufferedInputStream; import java.math.BigInteger; import java.util.ArrayList; import java.util.HashMap; import java.util.LinkedList; import java.util.List; import java.util.Map; import java.util.Queue; import java.util.Scanner; class Node { private static final int MAXP = 60; public BigInteger K; public long F; public int N; public int[] A; public Node() { K = BigInteger.ZERO; A = new int[MAXP]; } } public class Main { private static final int MAXIP = 250; private static final int MAXP = 60; private static BigInteger[] prime; private static void init() { boolean[] isPrime = new boolean[MAXIP]; for(int i=0;i<MAXIP;++i) { isPrime[i] = true; } isPrime[0] = isPrime[1] = false; for(int i=4;i<MAXIP;i+=2) { isPrime[i] = false; } for(int i=3;i<MAXIP;i+=2) { for(int j=3;i*j<MAXIP;j+=2) { isPrime[i*j] = false; } } prime = new BigInteger[MAXP]; for(int i=0, j=0;i<MAXIP;++i) { if(isPrime[i]) { prime[j++] = BigInteger.valueOf(i); } } } public static void main(String args[]) { init(); List<BigInteger> P = new ArrayList<BigInteger>(); BigInteger MP = BigInteger.ZERO; List<Node> ans = new ArrayList<Node>(); Scanner cin = new Scanner(new BufferedInputStream(System.in)); while(cin.hasNext()) { BigInteger temp = cin.nextBigInteger(); P.add(temp); if(temp.compareTo(MP) == 1) { MP = temp; } ans.add(new Node()); } Map<Long, BigInteger> map = new HashMap<Long, BigInteger>(); Queue<Node> queue = new LinkedList<Node>(); Node origin = new Node(); origin.K = BigInteger.ONE; origin.F = 1; origin.N = 0; queue.add(origin); map.put(origin.F, origin.K); while(!queue.isEmpty()) { Node u = queue.peek(); queue.remove(); BigInteger compare = map.get(u.F); if(compare != null) { if(compare.compareTo(u.K) == -1) { continue; } } for(int i=0;i<P.size();++i) { if(u.K.compareTo(P.get(i)) <= 0) { if(u.F > ans.get(i).F) { ans.get(i).F = u.F; ans.get(i).K = u.K; } else if(u.F == ans.get(i).F) { if(u.K.compareTo(ans.get(i).K) == -1) { ans.get(i).K = u.K; } } } } for(int i=0;i<u.N;++i) { Node v = new Node(); v.K = u.K.multiply(prime[i]); if(v.K.compareTo(MP) <= 0) { v.F = u.F / (u.A[i] + 1) * (u.A[i] + 2); v.N = u.N; for(int j=0;j<u.N;++j) { v.A[j] = u.A[j]; } ++ v.A[i]; boolean flag = true; compare = map.get(v.F); if(compare != null) { if(compare.compareTo(v.K) <= 0) { flag = false; } else { map.remove(v.F); } } if(flag) { queue.add(v); map.put(v.F, v.K); } } } Node v = new Node(); v.K = u.K.multiply(prime[u.N]); if(v.K.compareTo(MP) <= 0) { v.F = u.F * 2; v.N = u.N + 1; for(int i=0;i<u.N;++i) { v.A[i] = u.A[i]; } ++ v.A[u.N]; boolean flag = true; compare = map.get(v.F); if(compare != null) { if(compare.compareTo(v.K) <= 0) { flag = false; } else { map.remove(v.F); } } if(flag) { queue.add(v); map.put(v.F, v.K); } } } for(int i=0;i<ans.size();++i) { System.out.println(ans.get(i).K.toString() + " " + ans.get(i).F); //第一個(gè)數(shù)為滿足因子個(gè)數(shù)最多的K，第二個(gè)數(shù)為K的因子個(gè)數(shù)} } }

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