1083 List Grades

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

 

给定包含姓名、ID 和年级的 N 个学生记录的列表。你应该将记录按照成绩按非递增的顺序排序，并输出成绩在给定区间内的学生记录。

输入规范：每个输入文件包含一个测试用例。每个案例按以下格式给出：N name[1] ID[1] grade[1] name[2] ID[2] grade[2] ... ... name[N] ID[N] grade[ N] grade1 grade2 其中name[i]和ID[i]为不超过10个字符且无空格的字符串，grade[i]为[0, 100]范围内的整数，grade1和grade2为成绩区间的边界.保证所有等级都不同。

输出规范：对于每个测试用例，您应该输出成绩在给定区间 [grade1, grade2] 内且按非递增顺序排列的学生记录。每条学生记录占一行，学生姓名和 ID，以一个空格分隔。如果该区间内没有学生的成绩，则输出 NONE。

---

#include<iostream>
#include<string>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

struct _data
{
	string name;
	string id;
	int grade;
};

vector<_data>Stu;
vector<_data>nowStu;

bool cmp(_data a, _data b)
{
	return a.grade > b.grade;
}

int main()
{
	int n, m = 0;
	cin >> n;
	Stu.resize(n);
	for (int i = 0; i < n; i++)
	{
		cin >> Stu[i].name >> Stu[i].id >> Stu[i].grade;
	}
	int grade1, grade2;
	cin >> grade1 >> grade2;
	for (int i = 0; i < n; i++)
	{
		if (Stu[i].grade >= grade1 && Stu[i].grade <= grade2)

		{
			nowStu.push_back(Stu[i]);
			m++;
	
		}
	}

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

	if (nowStu.empty())
		cout << "NONE";
	else
	{
		for (int i = 0; i < m; i++)
		{
			cout << nowStu[i].name << " " << nowStu[i].id << endl;
		}
	}
	

	return 0;
}