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  • 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 }
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/13142092.html
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