PAT甲级1028

1028. List Sorting (25)

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

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, each contains a record of a student. A student's record consists of his or her distinct ID (a 6-digit number), name (a string with no more than 8 characters without space), and grade (an integer between 0 and 100, inclusive).

Output

For each test case, output the sorting result in N lines. That is, if C = 1 then the records must be sorted in increasing order according to ID's; if C = 2 then the records must be sorted in non-decreasing order according to names; and if C = 3 then the records must be sorted in non-decreasing order according to grades. If there are several students who have the same name or grade, they must be sorted according to their ID's in increasing order.

Sample Input 1
3 1
000007 James 85
000010 Amy 90
000001 Zoe 60
Sample Output 1
000001 Zoe 60
000007 James 85
000010 Amy 90
Sample Input 2
4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98
Sample Output 2
000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60
Sample Input 3
4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90
Sample Output 3
000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90

#include<cstdio>
#include<iostream>
#include<string.h>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct student
{
	char ID[20], name[20];
	int score;
};
bool cmp1(student s1, student s2)
{
	return strcmp(s1.ID, s2.ID)<=0;
}
bool cmp2(student s1, student s2)
{
	if (strcmp(s1.name, s2.name)!=0)
	{
		if (strcmp(s1.name, s2.name) < 0)
			return true;
		else
			return false;
	}
	else
		return strcmp(s1.ID, s2.ID) <= 0;//比较器返回的是true或false,必须这样写
}
bool cmp3(student s1, student s2)
{
	if (s1.score != s2.score)
		return s1.score < s2.score;
	else
		return strcmp(s1.ID, s2.ID) <= 0;
}
int main()
{
	int N, C;
	vector<student> v;
	student s;
	cin >> N >> C;
	for (int i = 0; i < N; i++)
	{
		//cin >> s.ID >> s.name >> s.score;
		scanf("%s %s %d", s.ID, s.name,&s.score);
		v.push_back(s);
	}
	switch (C)
	{
	case 1:sort(v.begin(), v.end(), cmp1); break;
	case 2:sort(v.begin(), v.end(), cmp2); break;
	case 3:sort(v.begin(), v.end(), cmp3); break;
	default:
		break;
	}
	for (int i = 0; i < v.size(); i++)
	{
		//cout << v[i].ID << " " << v[i].name << " " << v[i].score << endl;
		printf("%s %s %d\n", v[i].ID, v[i].name, v[i].score);
	}
	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