PAT_A1083#List Grades

概述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 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:

Nname[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:

4Tom CS000001 59Joe Math990112 89Mike CS991301 100Mary EE990830 9560 100

Sample Output 1:

Mike CS991301Mary EE990830Joe Math990112

Sample input 2:

2Jean AA980920 60Ann CS01 8090 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N10 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ID14 */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 node23 {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_JUDGE37 #else38     freopen("Test.txt","r",stdin);39 #endif40 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 }

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