1083 List Grades

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.

Input Specification:

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.

Output Specification:

For each test case you should output the student records of which the grades are in the given interval [grade1grade2] 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.

Sample Input 1:

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

Sample Output 1:

Mike CS991301
Mary EE990830
Joe Math990112
 

Sample Input 2:

2
Jean AA980920 60
Ann CS01 80
90 95
 

Sample Output 2:

NONE

题意:

  给出一个成绩单,按照成绩从大到小输出成绩在[grade1, grade2]之间的学生的信息。

思路:

  构造结构体,排序,输出。

Code:

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 struct Student {
 6     string name;
 7     string id;
 8     int grade;
 9 };
10 
11 bool cmp(Student a, Student b) { return a.grade > b.grade; }
12 
13 int main() {
14     int n;
15     cin >> n;
16     vector<Student> v;
17     for (int i = 0; i < n; ++i) {
18         string name, id;
19         int grade;
20         cin >> name >> id >> grade;
21         v.push_back({name, id, grade});
22     }
23     int grade1, grade2;
24     cin >> grade1 >> grade2;
25     bool found = false;
26     sort(v.begin(), v.end(), cmp);
27     for (int i = 0; i < n; ++i)
28         if (v[i].grade >= grade1 && v[i].grade <= grade2) {
29             found = true;
30             cout << v[i].name << " " << v[i].id << endl;
31         }
32     if (!found) cout << "NONE" << endl;
33 
34     return 0;
35 }
永远渴望,大智若愚(stay hungry, stay foolish)
原文地址:https://www.cnblogs.com/h-hkai/p/12797613.html