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:

 /*

 Data: 2019-07-13 10:38:02

 Problem: PAT_A1083#List Grades

 AC: 14:45

 题目大意：

 按成绩递减打印给定区间内学生的成绩

 输入：

 第一行给出，人数N

 接下来N行，姓名，ID，成绩

 最后一行给出，[g1,g2]

 输出：

 成绩递减，打印姓名和ID

 */

 #include<cstdio>

 #include<string>

 #include<vector>

 #include<iostream>

 #include<algorithm>

 const int M=1e3;

 using namespace std;

 struct node

 {

     string name,id;

     int grade;

 }info[M];

 vector<node> ans;

 bool cmp(node a, node b)

 {

     return a.grade > b.grade;

 }

 int main()

 {

 #ifdef    ONLINE_JUDGE

 #else

     freopen("Test.txt", "r", stdin);

 #endif

     int n,g1,g2;

     scanf("%d", &n);

     for(int i=; i<n; i++)

         cin >> info[i].name >> info[i].id >> info[i].grade;

     scanf("%d%d", &g1,&g2);

     for(int i=; i<n; i++)

         if(info[i].grade>=g1 && info[i].grade<=g2)

             ans.push_back(info[i]);

     sort(ans.begin(),ans.end(),cmp);

     if(ans.size() == )

         printf("NONE\n");

     for(int i=; i<ans.size(); i++)

         cout << ans[i].name << " " << ans[i].id << endl;

     return ;

 }

巴特西

PAT_A1083#List Grades

Source:

Description:

Input Specification:

Output Specification:

Sample Input 1:

Sample Output 1:

Sample Input 2:

Sample Output 2:

Keys:

Attention:

Code:

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