PAT甲级(Advanced Level)练习题 List Grades

题目描述

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.

输入描述:

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.

输出描述:

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.

输入例子:

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

输出例子:

Mike CS991301
Mary EE990830
Joe Math990112

题目分析及实现

本题实质是排序题，采用了选择排序，时间复杂度为O（n^2）,成功通过测试样例。

代码实现（C++版本）

#include <iostream>
using namespace std;
struct Student{
    string name;
    string id;
    int grade;
};
int main()
{
	//获得输入
    int num;
    cin >> num;
    Student stu[num];
    string ordername[num];
    string orderID[num];
    int ordergrade[num];
	
    for(int i =0; i < num;i++){
        cin >> stu[i].name >> stu[i].id >> stu[i].grade;
    }

    int minGrade,maxGrade;
    cin >> minGrade >>maxGrade;
	//选择排序
    for(int i =0;i<num;i++){
        int min = stu[i].grade;
        int temp = 0;
        for(int j = i+1; j<num;j++){
            if(stu[j].grade < min){
                min = stu[j].grade;
                temp = j;
            }
        }
        if(i<temp)
            swap(stu[i],stu[temp]);
    }
    //统计是否有NONE出现
    int count = 0;
    for(int i =0;i<num;i++){
        if(stu[i].grade>=minGrade && stu[i].grade<=maxGrade)
            count++;
    }
    if(count > 0){
    	//逆序打印
        for(int i =num-1;i>=0;i--){
            if(stu[i].grade>=minGrade && stu[i].grade<=maxGrade)
                cout<<stu[i].name << " "<<stu[i].id<<endl;
        }
    }else{
        cout<<"NONE"<<endl;
    }
    return 0;
}