[PAT A1083]List Grades

最新推荐文章于 2020-03-07 12:16:32 发布

ClemClementine

最新推荐文章于 2020-03-07 12:16:32 发布

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分类专栏： C++ 文章标签： PAT甲级

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本文详细解析PAT甲级ListGrades题目，指导如何按成绩区间和降序筛选并输出学生记录，附带C++实现代码。

[PAT A1083]List Grades

题目描述

1083 List Grades （25 分)Given a list of N student records with name, ID and grade. You are supposed to sort the records with respect to the grade in non-increasing order, and output those student records of which the grades are in a given interval.

输入格式

Each input file contains one test case. Each case is given in the following format:
N
name[1] ID[1] grade[1]
name[2] ID[2] grade[2]
… …
name[N] ID[N] grade[N]
grade1 grade2
where name[i] and ID[i] are strings of no more than 10 characters with no space, grade[i] is an integer in [0, 100], grade1 and grade2 are the boundaries of the grade’s interval. It is guaranteed that all the grades are distinct.

输出格式

For each test case you should output the student records of which the grades are in the given interval [grade1, grade2] and are in non-increasing order. Each student record occupies a line with the student’s name and ID, separated by one space. If there is no student’s grade in that interval, output NONE instead.

输入样例1

4
Tom CS000001 59
Joe Math990112 89
Mike CS991301 100
Mary EE990830 95
60 100

输出样例1

Mike CS991301
Mary EE990830
Joe Math990112

输入样例2

2
Jean AA980920 60
Ann CS01 80
90 95

输出样例2

NONE

解析

这道题通过率在PAT甲级里面为53%，是十分简单一道题目，如果不能一遍通过的话，可能就要好好反思了。PAT考试中25分的题目如果是这一道的话，那就真的是烧了高稥了。
这道题目一开始输入N，表示有N个学生，接下来N行，输入学生的姓名，id和他的成绩，然后输入grade1<grade2，我们要按照从高到低的顺序输出他们的姓名和id。
这道题目，没有两个人姓名相同，没有两个成绩是相同的，真的是从各方面实现了最简化，不多说只要是细心，一般都能直接过的，直接贴代码：

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct student
{
 string name, id;               
 int grade; 
};
bool cmp(student s1,student s2)
{
 return s1.grade > s2.grade;
}
int main()
{
 vector<student> stu;
 int N, min, max;
 cin >> N;
 for (int i = 0; i < N; i++) {
  student s;
  cin >> s.name >> s.id;
  scanf("%d", &s.grade);
  stu.push_back(s);
 }
 scanf("%d%d", &min, &max);
 sort(stu.begin(), stu.end(), cmp);
 int cnt = 0;
 for (int i = 0; i < stu.size(); i++) {
  if (stu[i].grade <= max && stu[i].grade >= min) {
   printf("%s %s\n", stu[i].name.c_str(), stu[i].id.c_str());
   cnt++;
  }
 }
 if (!cnt) printf("NONE\n");
 return 0;
}