PAT甲级1036

本题要求编程找出一组学生中女生最高分与男生最低分的学生信息,并计算两者分数之差。输入包括学生姓名、性别、ID及分数,输出需显示最高分女生与最低分男生的具体信息及分数差距。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

1036. Boys vs Girls (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N lines of student information. Each line contains a student's name, gender, ID and grade, separated by a space, where name and ID are strings of no more than 10 characters with no space, gender is either F (female) or M (male), and grade is an integer between 0 and 100. It is guaranteed that all the grades are distinct.

Output Specification:

For each test case, output in 3 lines. The first line gives the name and ID of the female student with the highest grade, and the second line gives that of the male student with the lowest grade. The third line gives the difference gradeF-gradeM. If one such kind of student is missing, output "Absent" in the corresponding line, and output "NA" in the third line instead.

Sample Input 1:

3
Joe M Math990112 89
Mike M CS991301 100
Mary F EE990830 95
Sample Output 1:
Mary EE990830
Joe Math990112
6
Sample Input 2:
1
Jean M AA980920 60
Sample Output 2:
Absent
Jean AA980920

NA

#include<stdio.h>
#include<iostream>
#include<string>
#include<map>
#include<vector>
#include<set>
#include<algorithm>
using namespace std;
struct stu
{
	string name, gender, ID;
	int grade;
	stu() 
		:name("Absent"), gender(""), ID(""), grade(-1) 
	{};//结构体初始化
	void operator=(stu a)
	{
		name = a.name;
		gender = a.gender;
		ID = a.ID;
		grade = a.grade;
	}
};
int main()
{
	int N;
	cin >> N; vector<stu> v; stu s;
	while (N--)
	{
		cin >> s.name >> s.gender >> s.ID >> s.grade;
		v.push_back(s);
	}
	int scoreMax = -1, scoreMin = 101;
	stu thebestgirl, theworstboy;
	for (int i = 0; i < v.size(); i++)
	{
		if (v[i].gender == "F")
		{
			if (scoreMax < v[i].grade)
			{
				scoreMax = v[i].grade;
				thebestgirl = v[i];
			}
		}
		if (v[i].gender == "M")
		{
			if (scoreMin >v[i].grade)
			{
				scoreMin = v[i].grade;
				theworstboy = v[i];
			}
		}
	}
	if (thebestgirl.name == "Absent")
		cout << "Absent" << endl;
	else
		cout << thebestgirl.name << " " << thebestgirl.ID << endl;
	if (theworstboy.name == "Absent")
		cout << "Absent" << endl;
	else
		cout << theworstboy.name << " " << theworstboy.ID << endl;
	if (thebestgirl.name == "Absent" || theworstboy.name == "Absent")
		cout << "NA";
	else
		cout << thebestgirl.grade - theworstboy.grade;

	return 0;
}


### 关于 PAT 甲级 1024 题目 PAT (Programming Ability Test) 是一项编程能力测试,其中甲级考试面向有一定编程基础的学生。对于 PAT 甲级 1024 题目,虽然具体题目描述未直接给出,但从相似类型的题目分析来看,这类题目通常涉及较为复杂的算法设计。 #### 数据结构的选择与实现 针对此类问题,常用的数据结构包括但不限于二叉树节点定义: ```cpp struct Node { int val; Node* lchild, *rchild; }; ``` 此数据结构用于表示二叉树中的节点[^1]。通过这种方式构建的二叉树能够支持多种遍历操作,如前序、中序和后序遍历等。 #### 算法思路 当处理涉及到图论的问题时,深度优先搜索(DFS)是一种常见的解题策略。特别是当需要寻找最优路径或访问尽可能多的节点时,结合贪心算法可以在某些情况下提供有效的解决方案[^2]。 #### 输入输出格式说明 根据以往的经验,在解决 PAT 类型的问题时,输入部分往往遵循特定模式。例如,给定 N 行输入来描述每个节点的信息,每行按照如下格式:“Address Data Next”,这有助于理解如何解析输入并建立相应的数据模型[^4]。 #### 数学运算示例 有时也会遇到基本算术表达式的求值问题,比如分数之间的加减乘除运算。下面是一些简单的例子展示不同情况下的计算结果: - \( \frac{2}{3} + (-2) = -\frac{7}{3}\) -2) = -\frac{4}{3}\) - \( \frac{2}{3} ÷ (-2) = -\frac{1}{3}\) 这些运算是基于样例提供的信息得出的结果[^3]。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值