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 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
大意:按给定信息将列表排序
输入:第一行:N为输入的数据(N小于等于100000),C为按第几列排序
后面N行:ID 姓名 分数
输出:将列表按第C列从小到大排序 如果出现C列有两个或者以上相等的情况则按ID号由小到大进行排序
思想:分三种情况定义3个比较函数
坑点:1.如果用C++ 中的std::cin>>和std::cout<<最后一个测试样例运行超时。
2.在main()函数中定义数组array[100000]可能 会爆栈。
知识点:在函数中定义的数组实在栈中 但系统栈是有限的,所以如果数组太大会爆掉,因此可以将他定义在函数之外作为全局变量,全局变量存储在静态存储区。(如有错误请留言,欢迎指正)。
#include<iostream>
#include<string.h>
#include<vector>
#include<algorithm>
#include<stdio.h>
using namespace std;
struct Student{
char id[10];
char name[10];
int grade;
};
Student student[100001];
void input(int n){
for(int i=0;i<n;i++){
scanf("%s%s%d",student[i].id,student[i].name,&student[i].grade);
}
}
int compare1(Student &a,Student &b){
return strcmp(a.id,b.id)<0;
}
int compare2(Student &a,Student &b){
if(strcmp(a.name,b.name))
return strcmp(a.name,b.name)<0;
return strcmp(a.id,b.id)<0;
}
int compare3(Student &a,Student &b){
if(a.grade!=b.grade)
return a.grade<b.grade;
return strcmp(a.id,b.id)<0;
}
int main(){
int n,c;
scanf("%d%d",&n,&c);
input(n);
switch(c){
case 1:sort(student,student+n,compare1);break;
case 2:sort(student,student+n,compare2);break;
case 3:sort(student,student+n,compare3);break;
}
for(unsigned int i=0;i<n;i++){
printf("%s %s %d\n",student[i].id,student[i].name,student[i].grade);
}
}