List Grades(PAT)

1.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.
给出N个学生的名字，ID和年级的记录列表。您应该按不递增的顺序对成绩进行排序，并输出成绩在给定间隔内的学生记录。

2.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.
每个输入文件包含一个测试用例。每个案例的格式如下：
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]是不超过10个字符的字符串，没有空格，grade[i]是[0,100]中的整数，grade1 grade2是级别间隔的边界。保证所有年级都是不同的。

3.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.
对于每个测试用例，您应该输出学生记录，其中的成绩在给定的间隔内[grade1, grade2]，并且是按不递增的顺序排列的。每个学生的记录与学生的姓名和ID占据一行，用一个空格隔开。如果在这段时间内没有学生的成绩，则输出“NONE”。

4.输入例子:

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

5.输出例子:

Mike CS991301
Mary EE990830
Joe Math990112

6.源代码：

#include<stdio.h>
typedef struct
{
	char name[10];
	char id[10];
	int grade;	
}student;

void qsort(student stu[],int L,int R)//快速排序
{
	if(L<R)
	{
		int i=L;
		int j=R;
		student key=stu[L];
		while(i<j)
		{
			while(i<j&&stu[j].grade<=key.grade)
				j--;
			if(i<j)
				stu[i++]=stu[j];
			while(i<j&&stu[i].grade>key.grade)
				i++;
			if(i<j)
				stu[j--]=stu[i];
		}
		stu[i]=key;
		qsort(stu,L,i-1);
		qsort(stu,i+1,R);
	}

}
int main()
{
	int i,count=0,N;
	//输入
	scanf("%d",&N);
	student stu[101];
	for(i=0;i<N;i++)
		scanf("%s %s %d",stu[i].name,stu[i].id,&stu[i].grade);
	int grade1,grade2;
	scanf("%d %d",&grade1,&grade2);
	//排序
	qsort(stu,0,N-1);
	//输出
	for(i=0;i<N;i++)
		if(stu[i].grade>=grade1&&stu[i].grade<=grade2)
		{
			count++;
			printf("%s %s\n",stu[i].name,stu[i].id);
		}
	//对符合条件的学生计数，如果没有就输出NONE
	if(count==0)
		printf("NONE\n");
	return 0;
}