PAT甲级 1141

1141 PAT Ranking of Institutions (25 分)

After each PAT, the PAT Center will announce the ranking of institutions based on their students' performances. Now you are asked to generate the ranklist.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤10​5​​), which is the number of testees. Then N lines follow, each gives the information of a testee in the following format:

ID Score School

where ID is a string of 6 characters with the first one representing the test level: B stands for the basic level, A the advanced level and T the top level; Score is an integer in [0, 100]; and School is the institution code which is a string of no more than 6 English letters (case insensitive). Note: it is guaranteed that ID is unique for each testee.

Output Specification:

For each case, first print in a line the total number of institutions. Then output the ranklist of institutions in nondecreasing order of their ranks in the following format:

Rank School TWS Ns

where Rank is the rank (start from 1) of the institution; School is the institution code (all in lower case); ; TWS is the total weighted score which is defined to be the integer part of ScoreB/1.5 + ScoreA + ScoreT*1.5, where ScoreX is the total score of the testees belong to this institution on level X; and Ns is the total number of testees who belong to this institution.

The institutions are ranked according to their TWS. If there is a tie, the institutions are supposed to have the same rank, and they shall be printed in ascending order of Ns. If there is still a tie, they shall be printed in alphabetical order of their codes.

Sample Input:

10
A57908 85 Au
B57908 54 LanX
A37487 60 au
T28374 67 CMU
T32486 24 hypu
A66734 92 cmu
B76378 71 AU
A47780 45 lanx
A72809 100 pku
A03274 45 hypu

Sample Output:

5
1 cmu 192 2
1 au 192 3
3 pku 100 1
4 hypu 81 2
4 lanx 81 2

题意 :一个学校派出多支队伍进行比赛,队伍分为A、B、T三级,将每个学校的多支队伍所得的分数按照给定公式进行计算,输出学校的排名、名称、得分以及参加比赛的队伍数量,若学校得分相同,则按照参加比赛的队伍数量进行升序排列,若参加比赛的队伍数量相同则按照学校名称根据字典序排列。

我的答案只得了20分,未AC~~~~~~

#include<iostream>

#include<algorithm>
#include<map>
#include<vector>
#include<unordered_map>
using namespace std;
struct node
{
	string school;
	int score;
	int num;
	node(){}
	node(string school1,int score1,int num1):school(school1),score(score1),num(num1){}
	friend  bool operator <(node A, node B) {
		if (A.score != B.score) return A.score > B.score;
		else if (A.num != B.num) return A.num < B.num;
		else if (A.school != B.school) return A.school <= B.school;
	}

};
unordered_map<string, int>mp;
unordered_map<string, int>m;
int main()
{
	int n;
	cin >> n;
	for (int i = 0;i < n;i++) {
		string s1, s2;
		int num;
		cin >> s1 >> num >> s2;
		for (int j = 0;s2[j];j++) {
			s2[j] = tolower(s2[j]);
		}
		if (s1[0] == 'A') mp[s2] += num;
		else if (s1[0] == 'B') mp[s2] += num / 1.5;
		else if (s1[0] == 'T') mp[s2] += num * 1.5;
		m[s2]++;
	}
	cout << mp.size() << endl;
	vector<node>s;
	for (unordered_map<string, int>::iterator it = mp.begin();it != mp.end();++it) {
		s.push_back(node(it->first, mp[it->first], m[it->first]));
	}
	sort(s.begin(), s.end());
	int flag = -1;

	for (int i = 0;i < s.size();i++) {
		if (flag == s[i].score) {
			cout << i << " " << s[i].school << " " << s[i].score << " " << s[i].num << endl;
			flag = -1;
		}
		else {
			cout << i +1<< " " << s[i].school << " " << s[i].score << " " << s[i].num << endl;
			flag = s[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]。
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