PAT A1083:List Grades之排序

题目描述

1083 List Grades (25分)
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

求解思路

按照策略进行排序即可

代码实现(AC)
#includecstdio
#includealgorithm
#includevector
using namespace std;
struct Student{
	char name[11];
	char id[11];
	int grade;
}stu[10001];
int n,grade1,grade2;
vectorStudent v;
bool cmp(Student a,Student b)
{
	return a.grade=b.grade;
}
void solve()
{
	scanf("%d",n);
	for(int i=0;in;i++)
	{
		scanf("%s %s %d",stu[i].name,stu[i].id,stu[i].grade);
	}
	scanf("%d %d",grade1,grade2);
	for(int i=0;in;i++)
	{
		if(stu[i].grade=grade1stu[i].grade=grade2)
			v.push_back(stu[i]);
	}
	if(v.size()==0)
	{
		printf("NONE");
	}
	else
	{
		sort(v.begin(),v.end(),cmp);
		for(int i=0;iv.size();i++)
			printf("%s %s\n",v[i].name,v[i].id);
	}
}
int main()
{
	solve();
	return 0;
}
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