【PAT】1028. List Sorting (25)
题目链接:http://pat.zju.edu.cn/contests/pat-a-practise/1028
题目描述:
Excel can sort records according to any column. Now you are supposed to imitate this function.
Input
Each input file contains one test case. For each case, the first line contains two integers N (<=100000) 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
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
分析:
输入学生信息,按照指定项进行排序。当按照name或grade排序时如果存在一样的情况就按照学号递增的形式输出。
原本是想用c++做,num和name用string类型表示。但是用c++中的cin进行输入时会有一组数据在最终提交时超时。所以改用scanf输入。
由于scanf不能输入string类型,所以将string类型用char[]代替。
但是如果num和name都用char[]类型,在输出时又会出错,见“错误代码一”。
后面就将num为int型,为了满足num为6位的要求,要用printf("%06d",num) 这种形式输出。
错误代码一:
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<vector>
#include<cstdio>
using namespace std; typedef struct student
{
char num[6];
char name[20];
int grade;
}Student; int n;
bool comparison(Student a,Student b)
{
if(n==1)
{
return strcmp(a.num,b.num)<0;
}
else if(n==2)
{
if(strcmp(a.name,b.name) == 0)
return strcmp(a.num,b.num);
return strcmp(a.name,b.name)<0;
}
else if(n==3)
{
if(a.grade == b.grade)
return strcmp(a.num,b.num) < 0;
return a.grade < b.grade;
}
} int main()
{
int M;
cin>>M>>n;
vector<Student> s(M);
int i;
for(i=0; i<M; i++)
//cin>>s[i].num>>s[i].name>>s[i].grade;
scanf("%s%s%d",&s[i].num,&s[i].name,&s[i].grade); sort(s.begin(),s.end(),comparison);
for(i=0; i<M; i++)
cout<<s[i].num<<" "<<s[i].name<<" "<<s[i].grade<<endl; return 0;
}
最终的输出不正确。但是
原因是???
正确代码:
参考:http://blog.csdn.net/sunbaigui/article/details/8657115
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<vector>
#include<cstdio>
using namespace std; typedef struct student
{
int num;
char name[20];
int grade;
}Student; int n;
bool comparison(Student a,Student b)
{
if(n==1)
{
return a.num<b.num;
}
else if(n==2)
{
if(strcmp(a.name,b.name) == 0)
return a.num<b.num;
return strcmp(a.name,b.name)<0;
}
else if(n==3)
{
if(a.grade == b.grade)
return a.num<b.num;
return a.grade < b.grade;
}
} int main()
{
int M;
cin>>M>>n;
vector<Student> s(M);
int i;
for(i=0; i<M; i++)
scanf("%d%s%d",&s[i].num,&s[i].name,&s[i].grade); sort(s.begin(),s.end(),comparison);
for(i=0; i<M; i++)
printf("%06d %s %d\n",s[i].num,s[i].name,s[i].grade); return 0;
}
【PAT】1028. List Sorting (25)的更多相关文章
- PAT 甲级 1028 List Sorting (25 分)(排序,简单题)
1028 List Sorting (25 分) Excel can sort records according to any column. Now you are supposed to i ...
- PAT 甲级 1028. List Sorting (25) 【结构体排序】
题目链接 https://www.patest.cn/contests/pat-a-practise/1028 思路 就按照 它的三种方式 设计 comp 函数 然后快排就好了 但是 如果用 c++ ...
- 【PAT甲级】1028 List Sorting (25 分)
题意: 输入一个正整数N(<=100000)和C(C属于{1,2,3}),接下来输入N行,每行包括学生的六位学号(习惯用string输入,因为可能有前导零),名字和成绩(正整数).输出排序后的信 ...
- 【PAT】1020 Tree Traversals (25)(25 分)
1020 Tree Traversals (25)(25 分) Suppose that all the keys in a binary tree are distinct positive int ...
- 【PAT】1015 德才论 (25)(25 分)
1015 德才论 (25)(25 分) 宋代史学家司马光在<资治通鉴>中有一段著名的“德才论”:“是故才德全尽谓之圣人,才德兼亡谓之愚人,德胜才谓之君子,才胜德谓之小人.凡取人之术,苟不得 ...
- 【PAT】1020. Tree Traversals (25)
Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and i ...
- 【PAT】B1055 集体照(25 分)
很简单的two points问题 ##注意:K是行数 #include<stdio.h> #include<string.h> #include<map> #inc ...
- 【PAT】1051 Pop Sequence (25)(25 分)
Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and p ...
- 【PAT】1063. Set Similarity (25) 待改进
Given two sets of integers, the similarity of the sets is defined to be Nc/Nt*100%, where Nc is the ...
随机推荐
- Linux Shall命令入门
Linux Shall命令入门 ifconfig //查看ip信息 service network start ...
- asp.net mvc4 简单使用Autofac依赖注入小结
1,首先使用 NuGet下载适当的Autofac版本 文件一,Autofac.3.5.2 文件二,Autofac.Mvc4.3.1.0 1,接口类 public interface IReposito ...
- GNU make 指南
http://docs.huihoo.com/gnu/linux/gmake.html GNU make 指南 翻译: 哈少 译者按: 本文是一篇介绍 GNU Make 的文章,读完后读者应该基本掌握 ...
- Druid 配置_StatFilter
Druid内置提供一个StatFilter,用于统计监控信息. 1. 别名配置 StatFilter的别名是stat,这个别名映射配置信息保存在druid-xxx.jar!/META-INF/drui ...
- erlang 大神
http://blog.csdn.net/erlib/article/details/46655905
- shell练习题
一.编写一个脚本使我们在写一个脚本时自动生成”#!/bin/bash”这一行和注释信息. 原文代码为: Shell 1 2 3 4 5 6 7 8 9 10 #!/bin/bash ...
- [翻译] JFDepthView 给view提供3D景深
JFDepthView 给view提供3D景深 https://github.com/atljeremy/JFDepthView This is an iOS project for presenti ...
- (原)将Oracle迁移到SQLServer
背景:中了一个标,Oracle改成SQLServer解决办法: 1.首先想到微软的解决方案:Microsoft SQL Server Migration Assistant v7.4 for Orac ...
- 解决Spring MVC报No converter found for return value of type:class java.util.ArrayList问题
一.背景 在搭建一套Spring+SpringMVC+Mybatis(SSM)的环境(搭建步骤会在以后博客中给出),结果运行 程序时,适用@ResponseBody注解进行返回List<对象&g ...
- sql数据库出现可疑
USE master GO SP_CONFIGURE 'allow updates',1 RECONFIGURE WITH OVERRIDE GO UPDATE SYSDATABASES SET ST ...