pat1083. List Grades (25)
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
#include<cstdio>
#include<stack>
#include<algorithm>
#include<iostream>
#include<stack>
#include<set>
#include<map>
using namespace std;
map<int,pair<string,string> > ha;//成绩-姓名-id
int main(){
//freopen("D:\\INPUT.txt","r",stdin);
int n;
scanf("%d",&n);
int i,g1,g2;
string name,id;
int grade;
for(i=;i<n;i++){
cin>>name>>id;
scanf("%d",&grade);
ha[grade]=make_pair(name,id);
}
int temp;
scanf("%d %d",&g1,&g2);
if(g1>g2){
temp=g1;
g1=g2;
g2=temp;
}
temp=;
map<int,pair<string,string> >::reverse_iterator it;
for(it=ha.rbegin();it!=ha.rend();it++){
if(it->first>=g1&&it->first<=g2){
cout<<(it->second).first<<" "<<(it->second).second<<endl;
temp++;
}
}
if(!temp){
printf("NONE\n");
}
return ;
}
pat1083. List Grades (25)的更多相关文章
- 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 ...
- 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 ...
- 【PAT甲级】1083 List Grades (25 分)
题意: 输入一个正整数N(<=101),接着输入N个学生的姓名,id和成绩.接着输入两个正整数X,Y(0<=X,Y<=100),逆序输出成绩在x,y之间的学生的姓名和id. tric ...
- 1083. List Grades (25)
the problem is from PAT,which website is http://pat.zju.edu.cn/contests/pat-a-practise/1083 and the ...
- PAT (Advanced Level) 1083. List Grades (25)
简单排序. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #i ...
- 1083. List Grades (25)-简单的排序
给定区间[L,R],给出在这区间之内的学生,并且按照他们的成绩非升序的顺序输出. #include <iostream> #include <cstdio> #include ...
- PAT_A1083#List Grades
Source: PAT A1083 List Grades (25 分) Description: Given a list of N student records with name, ID an ...
- PAT甲级题解(慢慢刷中)
博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给 ...
随机推荐
- C#设计模式(10)——组合模式
一.概念 组合模式有时候又叫做部分-整体模式,它使我们树型结构的问题中,模糊了简单元素和复杂元素的概念,客户程序可以向处理简单元素一样来处理复杂元素,从而使得客户程序与复杂元素的内部结构解耦. 二.组 ...
- centos6 启动流程
具体过程:1)加载BIOS的硬件信息,执行BIOS内置程序.2)读取MBR(Master Boot Record)中Boot Loader中的引导信息.3)加载内核Kernel boot到内存中.4) ...
- linux日常管理-系统服务
开机不启动不必要系统服务,节省硬件资源,解决安全隐患 调整服务有两种办法 没有这个命令就安装一下.执行这个命令,出现下面界面 按空格键选择或取消,tab切换选择确定或取消 系统服务留下 crond i ...
- Python学习:命令行运行,循环结构
一.安装配置和运行方法 1.安装OpenCV 3.1: 假设安装目录为"C:\Python34" 2.配置环境变量: 方法一:直接配置:打开"控制面板",搜索& ...
- IO系列之File
1 File类 1.1 目录列表器 在这里我主要是参考Think in Java的内容从而做的一些总结以及扩展.Java中的IO流的设计应该说是Java中最经典的,最学院式的设计,包括它的整体架构设计 ...
- Nodejs调试技术
基于Chrome浏览器的调试器 既然我们可以通过V8的调试插件来调试,那是否也可以借用Chrome浏览器的JavaScript调试器来调试呢?node-inspector模块提供了这样一种可能.我们需 ...
- 【总结整理】JQuery基础学习---DOM篇
前言: 先介绍下需要用到的浏览器提供的一些原生的方法(这里不处理低版本的IE兼容问题) 创建流程比较简单,大体如下: 创建节点(常见的:元素.属性和文本) 添加节点的一些属性 加入到文档中 流程中涉及 ...
- JS设置cookie、读取cookie、删除cookie(转)
JS设置cookie.读取cookie.删除cookie 转载 2015-04-17 投稿:hebedich 我要评论 Js操作Cookie总结(设置,读取,删除),工作中经常会用到的哦! ...
- (转)JAVA中的权限修饰符
注:本博文是转载的,原文地址:http://blog.csdn.net/xk632172748/article/details/51755438 Java中修饰符总结: 访问控制修饰符 访问控制修饰符 ...
- 2、Tophat align_summary.txt and samtools flagstat accepted_hits.bam disagree
###https://www.biostars.org/p/195758/ Left reads: Input : 49801387 Mapped : 46258301 (92.9% of input ...