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
- 调用bios喇叭发声
话不多说,上代码: #include <windows.h> #include <iostream> #include <map> using namespace ...
- 使用GDB命令行调试器调试C/C++程序
原文:http://xmodulo.com/gdb-command-line-debugger.html作者: Adrien Brochard 没有调试器的情况下编写程序时最糟糕的状况是什么?编译时跪 ...
- 程序员编程艺术:第三章续、Top K算法问题的实现
程序员编程艺术:第三章续.Top K算法问题的实现 作者:July,zhouzhenren,yansha. 致谢:微软100题实现组,狂想曲创作组. 时间:2011年05月08日 ...
- FPGrowth 实现
在关联规则挖掘领域最经典的算法法是Apriori,其致命的缺点是需要多次扫描事务数据库.于是人们提出了各种裁剪(prune)数据集的方法以减少I/O开支,韩嘉炜老师的FP-Tree算法就是其中非常高效 ...
- Leetcode_234_Palindrome Linked List
本文是在学习中的总结,欢迎转载但请注明出处:http://blog.csdn.net/pistolove/article/details/47334465 Given a singly linked ...
- OpenCV stereo matching 代码 matlab实现视差显示
转载请注明出处:http://blog.csdn.net/wangyaninglm/article/details/44151213, 来自:shiter编写程序的艺术 基础知识 计算机视觉是一门研究 ...
- sieve的objective-c实现
用obj-cl来实现前面的sieve代码貌似"丑"了不少,应该有更好的方式:比如不用Foundation或不用NSArray类,而改用其它更"底层"的类. 先把 ...
- Linux 系统应用编程——进程基础
一.Linux下多任务机制的介绍 Linux有一特性是多任务,多任务处理是指用户可以在同一时间内运行多个应用程序,每个正在执行的应用程序被称为一个任务. 多任务操作系统使用某种调度(shedule)策 ...
- JAVA加密技术-----MD5 与SHA 加密
关于JAVA的加密技术有很多很多,这里只介绍加密技术的两种 MD5与 SHA. MD5与SHA是单向加密算法,也就是说加密后不能解密. MD5 ---信息摘要算法,广泛用于加密与解密技术,常用于文件校 ...
- IOS常用第三方库《转》
UI 动画 网络相关 Model 其他 数据库 缓存处理 PDF 图像浏览及处理 摄像照相视频音频处理 响应式框架 消息相关 版本新API的Demo 代码安全与密码 测试及调试 AppleWatch ...