HDOJ1405 The Last Practice(有坑)

The Last Practice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11287    Accepted Submission(s): 2583

Problem Description

Tomorrow is contest day, Are you all ready?
We have been training for 45 days, and all guys must be tired.But , you are so lucky comparing with many excellent boys who have no chance to attend the Province-Final.

Now, your task is relaxing yourself and making the last practice. I guess that at least there are 2 problems which are easier than this problem.
what does this problem describe?
Give you a positive integer, please split it to some prime numbers, and you can got it through sample input and sample output.

 

Input

Input file contains multiple test case, each case consists of a positive integer n(1<n<65536), one per line. a negative terminates the input, and it should not to be processed.

 

Output

For each test case you should output its factor as sample output (prime factor must come forth ascending ), there is a blank line between outputs.

 

Sample Input

6012-1

 

Sample Output

Case 1.2 2 3 1 5 1Case 2.2 2 3 1
Hint
60=2^2*3^1*5^1
 

 这个题目最坑的莫过于空格问题，看题目是没说明的，看Sample Output也看不出来，第一次就PE了。

其次就是素数因子的问题，其实从2开始除，能除尽的都是素数了。

并没必要打表，我刚开始是打表的，被自己蠢哭了。

下面AC代码：

import java.util.Scanner;public class Main{private static Scanner scanner;public static void main(String[] args) {scanner = new Scanner(System.in);int cases = 1;while (scanner.hasNext()) {int n = scanner.nextInt();if (n < 0) {// 1<n<65536break;}if (cases > 1) {System.out.println();}System.out.println("Case " + cases + ".");cases++;for (int i = 2; i <= Math.sqrt(n); i++) {int count = 0;while (n % i == 0) {n /= i;count++;}if (count > 0) {System.out.print(i + " " + count + " ");}}if (n != 1) {System.out.print(n + " " + 1 + " ");}System.out.println();}}}

然后附上我第一次使用的方法（使用了集合，虽然AC了，但太复杂，仅供参考）：

import java.util.ArrayList;import java.util.Iterator;import java.util.List;import java.util.Map;import java.util.Map.Entry;import java.util.Scanner;import java.util.TreeMap;public class Main{private static Scanner scanner;private static List<Integer> list;public static void main(String[] args) {dabiao();scanner = new Scanner(System.in);int cases = 1;while (scanner.hasNext()) {int n = scanner.nextInt();if (n < 0) {break;}int s = n, t = 0;// key是素数，value是该素数对应的次方Map<Integer, Integer> map = new TreeMap<Integer, Integer>();while (s != 1) {for (int i = t; i < list.size(); i++) {int key = list.get(i);if (s % key == 0) {// System.out.println(key);if (map.containsKey(key)) {int value = map.get(key);map.replace(key, ++value);} else {map.put(key, 1);}t = i;s /= key;break;}}}Iterator<Entry<Integer, Integer>> iterator = map.entrySet().iterator();if (cases > 1) {System.out.println();}System.out.println("Case " + cases + ".");cases++;if (n == 1) {System.out.println(1 + " " + 1 + " ");}// boolean isFirst = true;while (iterator.hasNext()) {Map.Entry<Integer, Integer> entry = iterator.next();int key = entry.getKey();int value = entry.getValue();System.out.print(key + " " + value + " ");//后面都有个空格}System.out.println();}}// 65536private static void dabiao() {boolean isPrime[] = new boolean[65536];for (int i = 2; i < isPrime.length; i++) {isPrime[i] = true;}for (int i = 2; i < isPrime.length; i++) {if (isPrime[i]) {for (int j = i + i; j < isPrime.length; j += i) {isPrime[j] = false;}}}list = new ArrayList<Integer>();for (int i = 2; i < isPrime.length; i++) {if (isPrime[i]) {list.add(i);}}}}