List Sorting
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 (≤) 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
题目解析
本题给出两个整数，n为学生数量，c为排序指令，之后n行为学生信息，每个学生的信息包括学生号码ID，学生姓名name与学生成绩grade，排序指令为1时要求按照id升序排序，指令为2时要求按照学生姓名name字典序升序排序，若两个学生姓名字典序相同则按id升序排序，指令为3时按成绩grade升序排序，成绩相同时按id升序排序。
只需要将所有学生信息记录在一个容器vector中，根据题意对应的cmp函数，根据指令调用cmp函数即可。
AC代码
1 #include <bits/stdc++.h> 2 using namespace std; 3 struct student{ //结构体student代表一个学生的信息 4 int id; //学号 5 string name; //姓名 6 int grade; //成绩 7 }; 8 vector<student> V; //容器V记录所有学生信息 9 bool cmp1(student a, student b){ 10 //指令为1时按学号升序排序 11 return a.id < b.id; 12 } 13 bool cmp2(student a, student b){ 14 //指令为2时按姓名字典序升序排序 15 if(a.name != b.name) 16 return a.name < b.name; 17 else 18 //姓名字典序相同时按照学号升序排序 19 return a.id < b.id; 20 } 21 bool cmp3(student a, student b){ 22 //指令为3时按照成绩升序排序 23 if(a.grade != b.grade) 24 return a.grade < b.grade; 25 else 26 //成绩相同时按照学号升序排序 27 return a.id < b.id; 28 } 29 int n, c; 30 int main() 31 { 32 scanf("%d%d", &n, &c); //输入学生数量与排序指令 33 student temp; 34 for(int i = 0; i < n; i++){ 35 cin >> temp.id >> temp.name >> temp.grade; 36 //输入学生信息加入容器V 37 V.push_back(temp); 38 } 39 //根据排序指令进行排序 40 if(c == 1) 41 sort(V.begin(), V.end(), cmp1); 42 else if(c == 2) 43 sort(V.begin(), V.end(), cmp2); 44 else 45 sort(V.begin(), V.end(), cmp3); 46 //输出时使用cout会超时 47 for(auto i : V){ 48 printf("%06d %s %2d ", i.id, (i.name).c_str(), i.grade); 49 //格式化输出 50 } 51 return 0; 52 }