【hihocoder#1341】(微软探星夏令营在线技术笔试第一题)
来源:互联网 发布:驻马店公务员网络培训 编辑:程序博客网 时间:2024/05/17 08:23
第一次写这样的博客,毕竟很少在技术上追求一些东西,昨天微软探星夏令营的笔试被虐了,还是有不少感慨的,这里分享下第一题吧。
平时都是写工程里面简单的业务逻辑,很少做类似的oj题目,所以挺尴尬的只会一点点,代码很乱,我会适当解释一下哈。
本博客只是简单分享,顺便给自己9月份找工作攒人品,大神请无视我蹩脚的代码。
言归正传看题目:
#1341 : Constraint Checker
- 样例输入
2A<B<=E3<=E<52A 1B 2E 3A 3B 5E 10
- 样例输出
YesNo
描述
Given a set of constraints like 0<N<=M<=100 and values for all the variables, write a checker program to determine if the constraints are satisfied.
More precisely, the format of constraints is:
token op token op ... op token
where each token is either a constant integer or a variable represented by a capital letter and each op is either less-than ( < ) or less-than-or-equal-to ( <= ).
输入
The first line contains an integer N, the number of constraints. (1 ≤ N ≤ 20)
Each of the following N lines contains a constraint in the previous mentioned format.
Then follows an integer T, the number of assignments to check. (1 ≤ T ≤ 50)
Each assignment occupies K lines where K is the number of variables in the constraints.
Each line contains a capital letter and an integer, representing a variable and its value.
It is guaranteed that:
1. Every token in the constraints is either an integer from 0 to 1000000 or an variable represented by a capital letter from 'A' to 'Z'.
2. There is no space in the constraints.
3. In each assignment every variable appears exactly once and its value is from 0 to 1000000.
输出
For each assignment output Yes or No indicating if the constraints are satisfied.
就是一个约束匹配,第一排输入规则条数N(0-20),接下来根据条数N输入规则。
然后就是输入测试数据,先输入测试数据组数,然后输入测试数据的个数(这里需要注意,要根据规则里面字母出现的种类多少来输入)
解法和难点(个人的,有更好的欢迎留言或者贴地址):
1.利用规则,规则就是简单的< <= ,不难想到其实这就是一个不减的排序数据,去掉=(replace函数)然后利用<(split函数)来分割字符串就可以得到一个不减的排序数组了。
2.统计好规则中出现的字母种类数据,然后利用输入的数据替换掉字母得到N条规则得到的N排不减的数据
3.写一个简答的判断函数来判断这N排数据是否不减。
我的解法就是这样,主要还是对于oj题目不熟悉,输入输出的控制花了很多时间。
JAVA版本代码如下,写的非常粗糙,很多for循环,时间复杂度有点高,但是成功AC了,大家可以适当看看自己改写下:
import java.util.ArrayList;import java.util.Arrays;import java.util.List;import java.util.Scanner;public class Main { public void bubbleSort(int[] array) { int temp; for(int i=0;i<array.length;i++){ for(int j=0;j<array.length-i-1;j++){ if(array[j]>array[j+1]){ temp=array[j]; array[j]=array[j+1]; array[j+1]=temp; } } } } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); String[] rule = new String[n]; String rules = ""; for (int i = 0; i < n; i++) { String buffer = new String(scanner.next()); rule[i] = buffer; rules = rules +buffer; } rules = rules.replace("<=",""); rules = rules.replace("<",""); for (int i = 0; i <= 9; i++) { rules = rules.replace(Integer.toString(i),""); } List<String> data = new ArrayList<String>(); for (int i = 0; i < rules.length(); i++) { String s = rules.substring(i, i + 1); if (!data.contains(s)) { data.add(s); } } String result = ""; for (String s : data) { result += s; } int m = scanner.nextInt(); for (int i = 0; i < m; i++) { String[] zimu = new String[result.length()]; int[] shuzi = new int[result.length()]; boolean flag = true; for (int j = 0; j < result.length(); j++) { zimu[j] = scanner.next(); shuzi[j] = scanner.nextInt(); } for(int l=0;l<n;l++){ String temp = rule[l]; temp = temp.replace("=",""); for(int k=0;k<result.length();k++){ temp = temp.replace(zimu[k],Integer.toString(shuzi[k])); } String[] strArr = temp.split("<"); int[] numberArr = new int[strArr.length]; for (int s = 0; s <strArr.length; s++) { numberArr[s] = Integer.parseInt(strArr[s]); } Main ceshi = new Main(); int[] newnumberArr = new int[strArr.length]; for (int s = 0; s <strArr.length; s++) { newnumberArr[s] = Integer.parseInt(strArr[s]); } ceshi.bubbleSort(newnumberArr); if (!Arrays.equals(numberArr, newnumberArr)) { flag = false;} } if(flag) { System.out.println("Yes"); } else { System.out.println("No");} } }}
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