1028 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
思路:
模拟,注意采用cin输入的话最后一组数据会被卡到。
Code:
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 struct Stu { 6 int id; 7 char name[10]; 8 int grade; 9 } students[100005]; 10 11 bool cmp1(Stu a, Stu b) { return a.id < b.id; } 12 bool cmp2(Stu a, Stu b) { return strcmp(a.name, b.name) <= 0; } 13 bool cmp3(Stu a, Stu b) { return a.grade <= b.grade; } 14 15 int main() { 16 int n, c; 17 scanf("%d%d", &n, &c); 18 for (int i = 0; i < n; ++i) 19 scanf("%d%s%d", &students[i].id, students[i].name, &students[i].grade); 20 if (c == 1) 21 sort(students, students + n, cmp1); 22 else if (c == 2) 23 sort(students, students + n, cmp2); 24 else 25 sort(students, students + n, cmp3); 26 for (int i = 0; i < n; ++i) 27 cout << setw(6) << setfill('0') << students[i].id << " " 28 << students[i].name << " " << students[i].grade << endl; 29 return 0; 30 }