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. 应用Redis分布式锁解决重复通知的问题

    研究背景: 这几天被支付宝充值后通知所产生的重复处理问题搞得焦头烂额, 一周连续发生两次重复充钱的杯具, 发事故邮件发到想吐..为了挽回程序员的尊严, 我用了Redis的锁机制. 事故场景: 支付宝下 ...

  2. DirectX:Vector

    Tag DirectX下的博客主要用于记录DirectX的学习过程,主要参考<DirectX 12 3D 游戏实战开发>. Vector in DirectX Shader的编写离不开数学 ...

  3. Java代码质量检查checkstyle, pmd, cpd, p3c,findbugs, jacoco, sonarquebe以及和Jenkins集成

    概述 又搞一边质量扫描插件,之前做过一遍,然后后面各种忽略,然后就放弃了,所以,应该寻找一种方法,循序渐进的实施.本次将实施一个基本的打包扫描方案,包含 checkstyle 固定团队编码风格,固定命 ...

  4. Skywalking总结

    步骤四,完善Agent:你会发现,你在skywalking的Web监控页面看到的项目名称并非你原有的项目名称,而是一个默认的—— Your_ApplicationName.这是因为你还没有配置.打开/ ...

  5. GNU Makefile中的条件控制结构

    在常见的编程语言中,使用条件控制结构诸如if ... else if ... else...是很寻常的事情,那么在GNU Makefile中如何使用呢? ifeq ifneq 例如:foo.sh #! ...

  6. Linux : Nginx相关

    nginx安装参考链接: https://www.cnblogs.com/kaid/p/7640723.html 自定义编译目录: https://blog.csdn.net/ainuser/arti ...

  7. # .NET Core下操作Git,自动提交代码到

    .NET Core下操作Git,自动提交代码到 转自博客园(阿星Plus) .NET Core 3.0 预览版发布已经好些时日了,博客园也已将其用于生产环境中,可见 .NET Core 日趋成熟 回归 ...

  8. VS报错,Metadata file 'xxx.dll' could not be found

    错误提示“Metadata file 'xxx.dll' could not be found”步骤如下:1.右键单击解决方案,然后单击“属性”.2.单击左侧的配置.3.确保选中了它找不到的项目的“生 ...

  9. C# winform窗体简单保存界面控件参数到xml

    引用网上的 XMLHelper 地址 :https://www.cnblogs.com/chnboy/archive/2009/04/02/1427652.html 稍做修改 using System ...

  10. Python——XPath提取某个标签下所有文本

    /text()获取指定标签下的文本内容,//text()获取指定标签下的文本内容,包括子标签下的文本内容,比较简单的是利用字符串相加: room_infos = li.xpath('.//a[@cla ...