PAT_A1083#List Grades

Source:

PAT A1083 List Grades (25 分)

Description:

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 [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.

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

Keys:

  • 模拟题

Attention:

  • 先筛选，再排序，可以减少时间复杂度

Code:

 1 /*
 2 Data: 2019-07-13 10:38:02
 3 Problem: PAT_A1083#List Grades
 4 AC: 14:45
 5 
 6 题目大意：
 7 按成绩递减打印给定区间内学生的成绩
 8 输入：
 9 第一行给出，人数N
10 接下来N行，姓名，ID，成绩
11 最后一行给出，[g1,g2]
12 输出：
13 成绩递减，打印姓名和ID
14 */
15 #include<cstdio>
16 #include<string>
17 #include<vector>
18 #include<iostream>
19 #include<algorithm>
20 const int M=1e3;
21 using namespace std;
22 struct node
23 {
24     string name,id;
25     int grade;
26 }info[M];
27 vector<node> ans;
28 
29 bool cmp(node a, node b)
30 {
31     return a.grade > b.grade;
32 }
33 
34 int main()
35 {
36 #ifdef    ONLINE_JUDGE
37 #else
38     freopen("Test.txt", "r", stdin);
39 #endif
40 
41     int n,g1,g2;
42     scanf("%d", &n);
43     for(int i=0; i<n; i++)
44         cin >> info[i].name >> info[i].id >> info[i].grade;
45     scanf("%d%d", &g1,&g2);
46     for(int i=0; i<n; i++)
47         if(info[i].grade>=g1 && info[i].grade<=g2)
48             ans.push_back(info[i]);
49     sort(ans.begin(),ans.end(),cmp);
50     if(ans.size() == 0)
51         printf("NONE\n");
52     for(int i=0; i<ans.size(); i++)
53         cout << ans[i].name << " " << ans[i].id << endl;
54 
55     return 0;
56 }

 

posted @ 2019-07-13 10:57  林東雨  阅读(181)  评论(0)  编辑  收藏  举报