PAT1083:List Grades
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
要求按成绩降序输出给定成绩区间学生的信息。考虑到成绩不超过100而且唯一,可以用桶排序的思想来直接排序输出而不用比较。
1.用两种桶name[101]和ID[101]分别存放姓名和ID。
2.用成绩代表桶的下标,把对应成绩的学生信息放入桶中。如Jack CS00001 60 ---等价于---> name[60] = "Jack", ID[60] = "CS00001"。
3.根据区间范围输出不为空的桶里面的信息即可。如果范围内的桶都为空输出"NONE"。
代码
#include<iostream>
#include<vector>
#include<string>
using namespace std;
vector<string> name();
vector<string> ID(); int main()
{
int N;
while(cin >> N)
{
string n,id;
for(int i = ;i < N;i++)
{
int grade;
cin >> n >> id >> grade;
name[grade] = n;
ID[grade] = id;
}
int j,k;
bool isNone = true;
cin >> j >> k;
for(;k >= j;k--)
{
if(name[k] != "")
{
isNone = false;
cout << name[k] << " " << ID[k] << endl;
}
}
if(isNone)
cout <<"NONE" << endl;
ID.clear();
name.clear();
}
}
PAT1083:List Grades的更多相关文章
- pat1083. List Grades (25)
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 l ...
- PAT 甲级 1083 List Grades (25 分)
1083 List Grades (25 分) Given a list of N student records with name, ID and grade. You are supposed ...
- A1083. List Grades
Given a list of N student records with name, ID and grade. You are supposed to sort the records with ...
- Taking water into exams could boost grades 考试带瓶水可以提高成绩?
Takeing a bottle of water into the exam hall could help students boost their grades, researchers cla ...
- PAT 1083 List Grades[简单]
1083 List Grades (25 分) Given a list of N student records with name, ID and grade. You are supposed ...
- PAT 甲级 1083 List Grades
https://pintia.cn/problem-sets/994805342720868352/problems/994805383929905152 Given a list of N stud ...
- PAT 1083 List Grades
#include <cstdio> #include <cstdlib> using namespace std; class Stu { public: ]; ]; }; i ...
- A1083 List Grades (25)(25 分)
A1083 List Grades (25)(25 分) Given a list of N student records with name, ID and grade. You are supp ...
- A1083 List Grades (25 分)
Given a list of N student records with name, ID and grade. You are supposed to sort the records with ...
随机推荐
- Oracle EBS BOM模块常用表结构
表名: bom.bom_bill_of_materials 说明: BOM清单父项目 BILL_SEQUENCE_ID NUMBER 清单序号(关键字)ASSEMBLY_ITEM_ID NUMBE ...
- Java-ServletContextListener
/** * Implementations of this interface receive notifications about * changes to the servlet context ...
- HBase缓存的使用
hbase中的缓存分了两层:memstore和blockcache. 其中memstore供写使用,写请求会先写入memstore,regionserver会给每个region提供一个memstore ...
- HBase replication使用
hbase-0.90.0的一个重要改进是引入了replication机制,使它的数据完整性得到了进一步的保障.虽然这一功能还不太完善,但是今后必然会变得更加重要. hbase的replication机 ...
- Leetcode(59)-Count Primes
题目: Description: Count the number of prime numbers less than a non-negative number, n. 思路: 题意:求小于给定非 ...
- LeetCode - 二叉树的最大深度
自己解法,欢迎拍砖 给定一个二叉树,找出其最大深度. 二叉树的深度为根节点到最远叶子节点的最长路径上的节点数. 说明: 叶子节点是指没有子节点的节点. 示例:给定二叉树 [3,9,20,null,nu ...
- Loader转换器
一.简介 webpack本身只能处理js模块,Loader可以理解为模块和资源的转换器,它本身是一个函数,接受文件作为参数,返回转换的结果.因此,我们就能通过require来加载任何类型的模块和文件. ...
- Mybatis解决jdbc编程的问题
1.1.1 Mybatis解决jdbc编程的问题 1. 数据库链接创建.释放频繁造成系统资源浪费从而影响系统性能,如果使用数据库链接池可解决此问题. 解决:在SqlMapConfig.xml中配置 ...
- DBC的故事
1.DBC定义 DBC(data base CAN)是汽车ECU间进行CAN通讯的报文内容,有了它相互之间才能听懂. 2.DBC查看 DBC是文本文件,可以用记事本打开,一般都用CANdb++,可以更 ...
- Angular TypeScript开发环境集成jQuery扩展插件
集成步骤: 1.安装jquery极其扩展插件库ts定义文件 npm install jquery --save npm install --save-dev @types/jquery npm ins ...