1083 List Grades (25 point(s))

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本文介绍了一个简单的排序问题，要求根据学生的成绩将记录进行非递增排序，并筛选出成绩在特定区间内的学生记录。通过使用C++中的结构体和sort函数，可以轻松实现这一功能。

1083 List Grades (25 point(s))

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

简单排序题。对于没给范围的题，申请一个动态结构体数组来完成。

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
struct record {
	string name;
	string id;
	int grade;
	bool operator < (const record &other) const{
		return grade>other.grade;
	}
};

int main(void){
	int n,g1,g2;cin>>n;
	record *R=new record[n];
	for(int i=0;i<n;i++) cin>>R[i].name>>R[i].id>>R[i].grade; 
	sort(R,R+n);
	cin>>g1>>g2;
	bool flag = false;
	for(int i=0;i<n;i++){
		if(R[i].grade>=g1&&R[i].grade<=g2){
			flag = true; 
			cout<<R[i].name<<" "<<R[i].id<<endl; 
		}
	}
	if(!flag) cout<<"NONE"<<endl;
	return 0;
}