写完后又参考其他人的写法修改了一下：发现既然只有取得final exam成绩的学生才有最终成绩，那在42行到54行代码中，完全可以把最终成绩算出来（没有进行final exam的学生无最终成绩，不进入最终的排序）。

#include <algorithm>
#include <iostream>
#include <string>
#include <unordered_map>
#include <vector>struct stu {std::string id;int Gp = -1;int Gmid = -1;int Gfinal = -1;int g = -1;stu(std::string _id) : id(_id) {}
};int P, M, N, grade, cnt;
std::string id;
std::vector<stu> vec, result;
std::unordered_map<std::string, int> students;bool cmp(const stu &a, const stu &b) {return a.g != b.g ? a.g > b.g : a.id < b.id;
}int main() {std::cin >> P >> M >> N;cnt = 0;for (int i = 0; i < P; ++i) {std::cin >> id >> grade;if (grade >= 200) {students.insert({id, cnt++});vec.push_back(stu(id));vec[vec.size() - 1].Gp = grade;}}for (int i = 0; i < M; ++i) {std::cin >> id >> grade;if (students.find(id) != students.end()) {vec[students[id]].Gmid = grade;}}for (int i = 0; i < N; ++i) {std::cin >> id >> grade;if (students.find(id) != students.end()) {stu curr = vec[students[id]];curr.Gfinal = grade;if (curr.Gmid > grade) {curr.g = 0.4 * curr.Gmid + 0.6 * grade + 0.5;} else {curr.g = grade;}result.push_back(curr);}}sort(result.begin(), result.end(), cmp);for (int i = 0; i < result.size(); ++i) {if (result[i].g >= 60) {std::cout << result[i].id << " " << result[i].Gp << " " << result[i].Gmid<< " " << result[i].Gfinal << " " << result[i].g << std::endl;}}return 0;
}

题目如下：

For a student taking the online course "Data Structures" on China University MOOC (http://www.icourse163.org/), to be qualified for a certificate, he/she must first obtain no less than 200 points from the online programming assignments, and then receive a final grade no less than 60 out of 100. The final grade is calculated by G=(Gmid−term×40%+Gfinal×60%) if Gmid−term>Gfinal, or Gfinal will be taken as the final grade G. Here Gmid−term and Gfinal are the student's scores of the mid-term and the final exams, respectively.

The problem is that different exams have different grading sheets. Your job is to write a program to merge all the grading sheets into one.

Input Specification:

Each input file contains one test case. For each case, the first line gives three positive integers: P , the number of students having done the online programming assignments; M, the number of students on the mid-term list; and N, the number of students on the final exam list. All the numbers are no more than 10,000.

Then three blocks follow. The first block contains P online programming scores Gp's; the second one contains M mid-term scores Gmid−term's; and the last one contains N final exam scores Gfinal's. Each score occupies a line with the format: StudentID Score, where StudentID is a string of no more than 20 English letters and digits, and Score is a nonnegative integer (the maximum score of the online programming is 900, and that of the mid-term and final exams is 100).

Output Specification:

For each case, print the list of students who are qualified for certificates. Each student occupies a line with the format:

StudentID Gp Gmid−term Gfinal G

If some score does not exist, output "−1" instead. The output must be sorted in descending order of their final grades (G must be rounded up to an integer). If there is a tie, output in ascending order of their StudentID's. It is guaranteed that the StudentID's are all distinct, and there is at least one qullified student.

1137 Final Grading （PAT 甲级）相关推荐

PAT甲级1137 Final Grading：[C++题解]结构体、排序、哈希表、结构体构造函数、结构体内写函数
文章目录题目分析题目链接题目分析分析: 首先一个学生有id,另外有4个成绩:编程成绩.期中成绩.期末成绩.总评成绩.现有3个成绩单:编程成绩.期中成绩.期末成绩,让计算总评成绩,并排序输出. ...
PAT甲级1141 PAT Ranking of Institutions ：[C++题解]结构体、排序、哈希表、结构体构造函数、结构体内写函数、排名
文章目录题目分析题目来源题目分析来源:acwing 分析:和下面这题是一道题: PAT甲级1137 Final Grading:[C++题解]结构体.排序.哈希表.结构体构造函数.结构体内写函 ...
2020年9月PAT甲级满分必备刷题技巧
2020年7月的考试结束了,除了本次的考题更新,短期内不会更新. [7月题目的特点:首次线上考试,没出链表.树相关的模板题,第2到4题背景新颖,大大降低了抄袭历年代码的可能性,可以看作是线上考试的新趋 ...
【PAT甲级真题整理五】1121~1155
终于考完了qaq把最后一堆也整理出来了目录 1121 Damn Single(25)set.map的使用 1122 Hamiltonian Cycle(25)哈密顿回路 1123 Is It a C ...
刷PAT甲级的各题思路、细节以及遇到的问题记录
1001 A+B Format (20分) 因为一定会用到字符串,而string非常好用,但是用的时候一定要注意不能越界访问,否则会在运行时出现abort() has been called. 100 ...
PAT甲级题目翻译+答案 AcWing（哈希表）
1048 Find Coins (25 分) 题意 : 给一序列,要求从中选出两个值使得它们的和恰好等于m,输出第一个值字典序最小且升序的方案思路 : 看似O(n2)O(n^2)O(n2),其实可以 ...
PAT 甲级-算法初步
阅读原文接上一篇 PAT 甲级-入门模拟 ,自我感觉这部分才是真正的算法入门,对基础的数据结构提供了很好的类型题进行匹配练习包括分类:排序.散列.贪心.二分.双指针.打表.递推排序思想解释排 ...
【PAT甲级】A1101-A1155刷题记录
文章目录 (递推) A1101 Quick Sort (25 分) 0.23 (静态二叉树+遍历) A1102 Invert a Binary Tree (25 分) 0.51 (数学问题) A110 ...
PAT甲级 1056
PAT甲级 1056 题目 Mice and Rice 解析代码题目 Mice and Rice Mice and Rice is the name of a programming contes ...

1137 Final Grading （PAT 甲级）

题目如下：

Input Specification:

Output Specification:

1137 Final Grading （PAT 甲级）相关推荐

最新文章

热门文章