【Java】PAT乙级真题全记录（三）41到60题

最新推荐文章于 2022-12-26 16:51:08 发布

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最新推荐文章于 2022-12-26 16:51:08 发布

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分类专栏： PAT - Java 文章标签： PAT Java

本文链接：https://blog.youkuaiyun.com/qq_40624532/article/details/99630020

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这篇博客记录了PAT乙级考试中41到60题的解题思路和Java实现，涉及快速排序、划拳游戏、复数乘法等题目，分析了部分题目的坑点和效率要求，如测试点运行超时和答案格式问题。

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1041 考试座位号

import java.io.InputStreamReader;
import java.io.BufferedReader;
public class Main {
   
	public static void main(String[] args) throws Exception {
   
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine().trim());
		String[] info = new String[n + 1];    // n个考生的信息，为了便于查找初始化n+1个
		for(int i = 0; i < n; i++) {
   
			String[] temp = br.readLine().trim().split("\\s+");
			int testPos = Integer.parseInt(temp[1]); 	// 试机座位号
			info[testPos] = temp[0] + " " + temp[2];	// 试机座位号对应的其他信息
		}
		
		int m = Integer.parseInt(br.readLine().trim());
		int[] test = new int[m];						// m个待查询试机座位号
		String[] temp = br.readLine().trim().split("\\s+");
		for(int j = 0; j < m; j++) {
   
			test[j] = Integer.parseInt(temp[j]);
		}
		
		for(int j = 0; j < m; j++) {
   
			System.out.println(info[test[j]]);
		}
	}
}

1042 字符统计

import java.io.InputStreamReader;
import java.io.BufferedReader;
public class Main {
   
	public static void main(String[] args) throws Exception {
   
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String str = br.readLine().trim().toLowerCase();

		Character ch; // 当前字符
		int max = 0;  // 最大字频
		int[] count = new int[26]; // 字母字频统计
		for (int i = 0; i < str.length(); i++) {
   
			ch = str.charAt(i);
			if (Character.isLetter(ch)) {
   
				int index = ch - 'a';
				count[index]++;
				if (count[index] > max)
					max = count[index];
			}
		}
		
		/* 按照字母顺序输出第一个字频最大的 */
		for (int i = 0; i < 26; i++) {
   
			if (count[i] == max) {
   
				System.out.println((char) (i + 'a') + " " + max);
				break;
			}
		}
	}
}

1043 输出PATest

方法一：

import java.io.InputStreamReader;
import java.io.BufferedReader;
public class Main {
   	
	public static void main(String[] args) throws Exception {
   
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String str = br.readLine().trim();

		int[] count = new int[6]; // 对PATest的每个字母进行计数
		String sample = "PATest";
		for (int i = 0; i < str.length(); i++) {
   
			for (int j = 0; j < 6; j++) {
   
				if (str.charAt(i) == sample.charAt(j)) {
   
					count[j]++;
				}
			}
		}

		int total = count[0] + count[1] + count[2] + count[3] + count[4] + count[5];
		while (total > 0) {
    // 直到全部输出
			for (int i = 0; i < 6; i++) {
   
				if (count[i] > 0) {
   
					count[i]--;
					total--;
					System.out.print(sample.charAt(i));
				}
			}
		}
	}
}

方法二：

import java.io.InputStreamReader;
import java.util.HashMap;
import java.io.BufferedReader;
public class Main {
   
	public static void main(String[] args) throws Exception {
   
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String str = br.readLine().trim();
		String sample = "PATest";

		HashMap<Character, Integer> count = new HashMap<>();
		for (int i = 0; i < 6; i++) {
   
			count.put(sample.charAt(i), 0); // 初始化
		}

		for (int i = 0; i < str.length(); i++) {
   
			count.computeIfPresent(str.charAt(i), (k, v) -> v + 1); // 计数
		}

		Character key;
		int total = count.get('P') + count.get('A') + count.get('T') + count.get('e') + count.get('s') + count.get('t');
		while (total > 0) {
   
			for (int i = 0; i < 6; i++) {
   
				key = sample.charAt(i);
				if (count.get(key) > 0) {
   
					System.out.print(key);
					total--;
				}
				count.computeIfPresent(key, (k, v) -> v - 1); // 输出
			}
		}
	}
}

1044 火星数字

唯一的坑就是！13、26这样的十进制，写成十三进制是10、20，但是火星文是tam、hel，也就是说，个位的0不需要表示出来。

import java.io.InputStreamReader;
import java.io.BufferedReader;
public class Main {
   
	static String[] sample1 = {
    "tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec" }; // 低位数字
	static String[] sample2 = {
    "tret", "tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou" }; // 高位数字

	public static void main(String[] args) throws Exception {
   
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine().trim());
		String[] words = new String[n];
		String[] trans = new String[n];

		for (int i = 0; i < n; i++) {
   
			words[i] = br.readLine().trim();
			if (Character.isLetter(words[i].charAt(0))) {
    // true则是火星文
				trans[i] = toEarth(words[i]);
			} else {
   
				trans[i] = toMartian(words[i]);
			}
		}

		for (int i = 0; i < n; i++)             // 输出结果
			System.out.println(trans[i]);
	}

	public static String toMartian(String s) {
    // 数字转火星文
		StringBuilder mars = new StringBuilder("");
		int earth = Integer.parseInt(s);

		if (earth == 0) {
   
			mars.append("tret");
		} else {
   
			int temp1 = earth % 13; // 低位
			int temp2 = (earth / 13) % 13; // 高位

			if (temp2 > 0)
				mars.append(sample2[temp2]);
			if(temp1 != 0 && temp2 != 0)
				mars.append(" ");
			if (temp1 > 0) // 低位为零时不输出
				mars.append(sample1[temp1]);
		}
		return mars.toString();
	}

	public static String toEarth(String s) {
    // 火星文转数字
		int earth = 0;
		String[] temp = s.split("\\s+");
		if (temp[0] != "tret") {
   
			for (int i = 0; i < temp.length; i++) {
   
				for (int j = 1; j < 13; j++) {
   
					if (temp[i].equals(sample1[j]))
						earth += j;
					if (temp[i].equals(sample2[j]))
						earth += j * 13;
				}
			}
		}
		return String.valueOf(earth);
	}
}

1045 快速排序（1、3、4、5测试点运行超时）

本题java劝退，典型的考效率的，时间复杂度不能超过O(n)，解决问题本身很简单。另外，坑点注意！测试点2是主元为0的情况，要输出 “0\n\n” 格式才正确。
思路一（较慢，只能过测试点0）：

import java.io.InputStreamReader;
import java.io.BufferedReader;

public class Main {
   

	public static void main(String[] args) throws Exception {
   
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.valueOf(br.readLine().trim());
		String[] temp =</