PAT 1083. List Grades (25)

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

思路：快速排序

#include <stdio.h>
#include <algorithm>
using namespace std;
struct Node
{
	char name[15], ID[15];
	int grade;
	bool friend operator < (Node n1, Node n2)
	{
		return n1.grade > n2.grade;
	}
};
Node list[10000];
int N,s,e;
int main()
{
	while(scanf("%d", &N) != EOF)
	{
		for(int i = 0; i < N; i++)
		{
			scanf("%s %s %d", &list[i].name, &list[i].ID, &list[i].grade);
		}
		scanf("%d %d", &s, &e);
		sort(list, list + N);
		int cnt = 0;
		for(int i = 0; i < N; i++)
		{
			if(list[i].grade <= e && list[i].grade >= s)
			{
				printf("%s %s\n", list[i].name, list[i].ID);
				cnt++;
			}
		}
		if(cnt == 0)
		{
			printf("NONE\n");
		}
	}
	return 0;
}