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
 #include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
typedef struct{
char name[];
char id[];
int grade;
}info;
bool cmp(info a, info b){
return a.grade > b.grade;
}
info stu[];
int main(){
int N, high, low, cnt = ;
scanf("%d", &N);
for(int i = ; i < N; i++)
scanf("%s %s %d", stu[i].name, stu[i].id, &(stu[i].grade));
sort(stu, stu + N, cmp);
scanf("%d%d", &low, &high);
for(int i = ; i < N; i++){
if(stu[i].grade >= low && stu[i].grade <= high){
printf("%s %s\n", stu[i].name, stu[i].id);
cnt++;
}
}
if(cnt == )
printf("NONE");
cin >> N;
return ;
}

A1083. List Grades的更多相关文章

  1. 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 ...

  2. 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 ...

  3. PAT甲级——A1083 List Grades

    Given a list of N student records with name, ID and grade. You are supposed to sort the records with ...

  4. PAT_A1083#List Grades

    Source: PAT A1083 List Grades (25 分) Description: Given a list of N student records with name, ID an ...

  5. PAT甲级题解分类byZlc

    专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d& ...

  6. PAT1083:List Grades

    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 ...

  7. 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 ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. 5分钟入门自动化测试——你应该学会的Postman用法(2)

    前言 之前的一篇文章<你应该学会的Postman用法>,主要介绍了postman的一些高级的用法,便于日常开发和调试使用,本文的基础是对postman的基本使用以及一些高级用法有一定的了解 ...

  2. java使用何种类型表示精确的小数?

    问题 java使用何种类型表示精确的小数? 结论 float和double类型的主要设计目标是为了科学计算和工程计算,速度快,存在精度丢失 BigDecimal用来表示任意精确浮点数运算的类,在商业应 ...

  3. 百度之星-day2-1004-二分答案

    由于序列有序,求其中一个最优解,二分答案即可,注意二分时上边界满足才保存 #include<iostream> #include<stdio.h> #include<st ...

  4. 《Linux内核设计与实现》读书笔记三

    Chapter 18 调 试 18.1 准备开始 1.准备工作: 一个bug 一个藏匿bug的内核版本 相关内核代码的知识和运气 2.执行foo就会让程序立即产生核心信息转储(dump core). ...

  5. 网络:Session原理及存储

    一.Session的工作流程 二.会话保持 会话保持是负载均衡最常见的问题之一,会话保持是指在负载均衡器上实现的一种机制,可以识别客户端与服务器之间交互过程的关连性,在作负载均衡的同时还保证一系列相关 ...

  6. C#-ToString格式化

    Int.ToString(format): 格式字符串采用以下形式:Axx,其中 A 为格式说明符,指定格式化类型,xx 为精度说明符,控制格式化输出的有效位数或小数位数,具体如下: 格式说明符 说明 ...

  7. Docker查看容器IP

    https://segmentfault.com/q/1010000001637726 https://blog.csdn.net/sannerlittle/article/details/77063 ...

  8. 【转】Linux tail 命令详解

    Linux tail 命令详解 http://www.2cto.com/os/201111/110143.html

  9. mysql 备份数据库 mysqldump

    @echo off for /F "usebackq tokens=1,2 delims==" %%i in (`wmic os get LocalDateTime /VALUE ...

  10. 【工具技巧】:sublime notepad++ 多行编辑

    1. 多行编辑 sublime 最简单的多行编辑实现方法 1. 鼠标选中文件 然后按 ctrl+D 自动选中相同的进行同时编辑 2.选中shift按键+鼠标右键进行选择,可以同时选中多行进行编辑. n ...