Excel can sort records according to any column. Now you are supposed to imitate this function.

Input Specification:

Each input file contains one test case. For each case, the first line contains two integers N (≤10

​5

​​ ) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, each contains a record of a student. A student's record consists of his or her distinct ID (a 6-digit number), name (a string with no more than 8 characters without space), and grade (an integer between 0 and 100, inclusive).

Output Specification:

For each test case, output the sorting result in N lines. That is, if C = 1 then the records must be sorted in increasing order according to ID's; if C = 2 then the records must be sorted in non-decreasing order according to names; and if C = 3 then the records must be sorted in non-decreasing order according to grades. If there are several students who have the same name or grade, they must be sorted according to their ID's in increasing order.

Sample Input 1:

3 1

000007 James 85

000010 Amy 90

000001 Zoe 60

Sample Output 1:

000001 Zoe 60

000007 James 85

000010 Amy 90

Sample Input 2:

4 2

000007 James 85

000010 Amy 90

000001 Zoe 60

000002 James 98

Sample Output 2:

000010 Amy 90

000002 James 98

000007 James 85

000001 Zoe 60

Sample Input 3:

4 3

000007 James 85

000010 Amy 90

000001 Zoe 60

000002 James 90

Sample Output 3:

000001 Zoe 60

000007 James 85

000002 James 90

000010 Amy 90

开始技术总结:

  • 主要就是cmp函数的编写,可以通过编写三个cmp函数来分别对应于三种排序的情况,然后根据输入得到的C值来选择哪个cmp函数应用,我使用的是通过定义一个全局变量C,然后编写一个cmp函数,在cmp函数里面区别三种排序情况。注意如果要使用C来判断,必须把C定义为全局变量。这个函数的参考代码如下:
bool cmp(Student a, Student b) {
if (C == 1) {
return a.id < b.id;
}
else if (C == 2) {
if (strcmp(a.name, b.name) == 0) {
return a.id < b.id;
}
else {
return strcmp(a.name, b.name) < 0;
} }
else if (C == 3) {
if (a.grade == b.grade) {
return a.id < b.id;
}
else {
return a.grade < b.grade;
}
}
}

所有部分的代码:

//在VS2017上编写,通过PAT测试满分25
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std; struct Student {
int id;
char name[10];
int grade;
}stu[100010]; int C;//全局变量用来在cmp函数使用,判定排序方式 bool cmp(Student a, Student b) {
if (C == 1) {
return a.id < b.id;
}
else if (C == 2) {
if (strcmp(a.name, b.name) == 0) {
return a.id < b.id;
}
else {
return strcmp(a.name, b.name) < 0;
} }
else if (C == 3) {
if (a.grade == b.grade) {
return a.id < b.id;
}
else {
return a.grade < b.grade;
}
}
}
int main() {
int N;//分别记录人数N
scanf("%d%d", &N, &C);
for (int i = 0; i < N; i++) {
scanf("%d%s%d", &stu[i].id, stu[i].name, &stu[i].grade);
}
sort(stu, stu + N, cmp);
for (int i = 0; i < N; i++) {
printf("%06d %s %d\n", stu[i].id, stu[i].name, stu[i].grade);
} system("pause");
return 0;
}

PATA1028 List Sorting的更多相关文章

  1. HDU Cow Sorting (树状数组)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2838 Cow Sorting Problem Description Sherlock's N (1  ...

  2. 1306. Sorting Algorithm 2016 12 30

    1306. Sorting Algorithm Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description One of the f ...

  3. 算法:POJ1007 DNA sorting

    这题比较简单,重点应该在如何减少循环次数. package practice; import java.io.BufferedInputStream; import java.util.Map; im ...

  4. U3D sorting layer, sort order, order in layer, layer深入辨析

    1,layer是对游戏中所有物体的分类别划分,如UIlayer, waterlayer, 3DModelLayer, smallAssetsLayer, effectLayer等.将不同类的物体划分到 ...

  5. WebGrid with filtering, paging and sorting 【转】

    WebGrid with filtering, paging and sorting by Jose M. Aguilar on April 24, 2012 in Web Development A ...

  6. ASP.NET MVC WebGrid – Performing true AJAX pagination and sorting 【转】

    ASP.NET MVC WebGrid – Performing true AJAX pagination and sorting FEBRUARY 27, 2012 14 COMMENTS WebG ...

  7. poj 1007:DNA Sorting(水题,字符串逆序数排序)

    DNA Sorting Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 80832   Accepted: 32533 Des ...

  8. ural 1252. Sorting the Tombstones

    1252. Sorting the Tombstones Time limit: 1.0 secondMemory limit: 64 MB There is time to throw stones ...

  9. CF#335 Sorting Railway Cars

    Sorting Railway Cars time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

随机推荐

  1. nginx报错111: Connection refused

    最近遇到了nginx疯狂抛错,access.log一天一共5W多条,但error.log中有大概9K多条,基本都是111: Connection refused,这到底是为什么呢? 从日志看起 我们还 ...

  2. Java学习:File类中的过滤器接口

    javaIO类的File类应用:过滤器接口 FilenameFilter和FileFilter都是用来过滤文件的 例如: 过滤以.jpg或者.java结尾的文件. 通过看他们的源码: 通过使用File ...

  3. Java并发专栏(一)—— Process vs Thread

    一.前言 程序是代码和数据的集合,是一种静态实体.不具有代码执行和数据处理的能力,更多是一种行为的描述. 如果将程序和处理器结合,处理器将程序加载至内存,然后执行程序代码处理数据.这时就是可执行的程序 ...

  4. 《 .NET并发编程实战》阅读指南 - 第12章

    先发表生成URL以印在书里面.等书籍正式出版销售后会公开内容.

  5. Java 函数式编程--流操作

    GitHub Page: http://blog.cloudli.top/posts/Java-函数式编程-流操作/ 外部迭代到内部迭代 在使用集合类时,通用的方式是在使用 for 循环集合上进行迭代 ...

  6. WPF 精修篇 动态资源

    原文:WPF 精修篇 动态资源 动态资源 使用 DynamicResource 关键字 静态 就是 StaticResource 原则上是 能用静态就用静态 动态会让前台界面压力很大~ 动态资源引用 ...

  7. 通过SharpZipLib实现文件夹压缩以及解压

    代码说明 基于SharpZipLib实现Zip压缩解压,扩展实现文件夹级别压缩解压: 项目源码:MasterChief.DotNet.Infrastructure.Zip Install-Packag ...

  8. 两道JVM面试题,竟让我回忆起了中学时代!

    作者:肥朝 原文链接:https://mp.weixin.qq.com/s/4wJ6ANal0blLOseasfIuVw 中学授课模式 考虑到可能有部分粉丝对JVM参数不清楚,所以我们参照中学的授课模 ...

  9. Flask模板渲染

    目录 Flask模板渲染 Jinja2模板引擎简介 模板 Jinja2 模板变量 变量 控制结构 宏,类似Python代码中的函数 模板继承 包含(Include) 过滤器 链式调用 常见内建过滤器 ...

  10. Robotframework ride ,运行后提示, [WinError 2] 系统找不到指定的文件。

    运行后提示, [WinError 2] 系统找不到指定的文件. command: pybot.bat --argumentfile C:\Users\123\AppData\Local\Temp\RI ...