PAT (Advanced Level) Practise 1083 List Grades (25)

最新推荐文章于 2018-01-15 09:17:39 发布

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最新推荐文章于 2018-01-15 09:17:39 发布

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本文介绍了一种算法，用于根据给定的分数区间对学生记录进行排序和筛选。输入包括学生的姓名、ID和成绩，输出为指定分数范围内的学生记录，并按成绩降序排列。

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1083. List Grades (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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：给出n个人的名字，id和分数，问在一个分数段里的有哪些人，按字典序输出

#include <iostream>    
#include <cstdio>    
#include <string>    
#include <cstring>    
#include <algorithm>    
#include <cmath>    
#include <vector>    
#include <map>    
#include <set>    
#include <queue>    
#include <stack>    
#include <functional>    
#include <climits>    

using namespace std;

#define LL long long    
const int INF = 0x7FFFFFFF;

char name[100009][25],id[100009][25];
int score[100009];
int n, l, r;
vector<int> ans;

bool cmp(const int &x, const int&y)
{
	return score[x] > score[y];
}

int main()
{
	while (~scanf("%d", &n))
	{
		ans.clear();
		for (int i = 0; i < n; i++) scanf("%s%s%d", name[i], id[i], &score[i]);
		scanf("%d%d", &l, &r);
		for (int i = 0; i < n; i++)
			if (score[i] >= l&&score[i] <= r) ans.push_back(i);
		sort(ans.begin(), ans.end(), cmp);
		if (ans.size())
		{
			for (int i = 0; i < ans.size(); i++)
				printf("%s %s\n", name[ans[i]], id[ans[i]]);
		}
		else printf("NONE\n");
	}
	return 0;
}