PAT A1083 List Grades (25分)-优快云博客

本文链接：https://blog.youkuaiyun.com/PanYiAn9/article/details/104280414

本文介绍了一种算法，用于对学生记录进行排序并筛选出特定分数区间内的学生信息。通过使用结构体数组和自定义比较函数，可以实现对学生记录的高效处理。

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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

题意：

输入学生总数，然后逐行输入学生姓名、id、分数，再输入允许输出的最低分和最高分；输出分数在最低分和最高分之间的学生的姓名和id，如果不存在这样的学生则输出NONE.(题目没有指定N的范围；题目保证每个学生的分数都不相同.)

思路：

(1)开设一个student结构体数组stu[]，包含name、id和grade；
(2)先对所有学生进行排序，再遍历整个数组进行判断，对分数满足条件的学生输出.

代码：

#include <cstdio>
#include <algorithm>
using namespace std;

struct student{
	char name[15],id[15];
	int grade;
}stu[100010];
bool cmp(student a,student b){
	return a.grade>b.grade;
}

int main(){
	int N;
	scanf("%d",&N);
	for(int 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);
	sort(stu,stu+N,cmp); 
	int input = false;//记录是否存在分数在最低分和最高分之间的学生 
	for(int i=0;i<N;i++){
		if(stu[i].grade>=grade1&&stu[i].grade<=grade2){
			printf("%s %s\n",stu[i].name,stu[i].id);
			input = true;
		}
	}
	if(input==false) printf("NONE\n");
	return 0;
}