PAT A1028 List Sorting (25 分)——排序,字符串输出用printf
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 (≤105) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then Nlines 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
#include <stdio.h>
#include <algorithm>
#include <string>
#include <iostream>
using namespace std;
const int maxn=;
struct stu{
string id;
string name;
int grade;
}students[maxn];
bool cmp1(stu s1,stu s2){
return s1.id<s2.id;
}
bool cmp2(stu s1,stu s2){
if(s1.name==s2.name) return s1.id<s2.id;
else return s1.name<s2.name;
}
bool cmp3(stu s1,stu s2){
if(s1.grade==s2.grade) return s1.id<s2.id;
else return s1.grade<s2.grade;
}
int main(){
int n,c;
scanf("%d %d",&n,&c);
for(int i=;i<n;i++){
cin>>students[i].id>>students[i].name>>students[i].grade;
}
if(c==){
sort(students,students+n,cmp1);
}
else if(c==){
sort(students,students+n,cmp2);
}
else if(c==){
sort(students,students+n,cmp3);
}
for(int i=;i<n;i++){
printf("%s %s %d\n",students[i].id.c_str(),students[i].name.c_str(),students[i].grade);
}
}
注意点:很简单的排序题,但是最后一个测试点会超时,原因是cout很慢,一开始以为是sort很慢,想是不是要用priority_queue,发现一样超时。然后把id改成int试了试还是超时,百度了一下,发现人家也都是sort做的唯一不同输出name也用printf。最后果然是cout速度太慢,printf输出string要用c_str。
最新文章
- 在PC上测试移动端网站和模拟手机浏览器的5大方
- 在其他系统Iframe中显示SharePoint 页面
- qt5.4.0编译错误
- 加密配置文件(App.Config和Web.config)中connectionStrings通用方法
- jsoncpp初使用
- Integer &; int &; == &; equals
- Tomcat启动后快逸报表报错的解决方法
- 如何将cmd中命令输出保存为TXT文本文件
- webView、scrollView、TableView,为了防止滚动时出现偏移,底部黑框问题等
- Mysql-学习笔记(==》存储过程 九)
- PHPWord生成word实现table合并(colspan和rowspan)
- fiddle 中 显示serverIp
- Chef
- requests设置headers,proxies,cookies
- R语言编程艺术# 数据类型向量(vector)
- 配置mybatis错误总结
- longest incresing sequence
- iOS 通知的使用
- CentOS/RHEL 7中的firewall控制
- 【Qt编程】基于Qt的词典开发系列<;五>;--无边框窗口的拖动