PAT(A) 1028. List Sorting (25)

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) 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

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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100005;

struct Node{
    int id;         //准考证号
    char name[10];  //姓名
    int score;      //分数
}stu[maxn];

//划重点:按三种排序规则写三个cmp(), 根据读入C的不同选择不同的排序函数
//C==1: 按准考证从小到大排序
bool cmp1(Node a, Node b){
    return a.id<b.id;           //按准考证从小到大排序
}
//C==2:
bool cmp2(Node a, Node b){
    int s=strcmp(a.name, b.name);
    if(s!=0) return s<0;        //按姓名字典序从大到小排序
    else     return a.id<b.id;  //姓名相同则按准考证号从小到大排序
}
//C==3:
bool cmp3(Node a, Node b){
    if(a.score!=b.score)    return a.score<b.score; //按分数从小到大排序
    else                    return a.id<b.id;       //分数相同则按准考证号从小到大排序
}

int main()
{
    int n, c;
    scanf("%d%d", &n, &c);
    for(int i=0; i<n; i++){
         //注意(1): 用%s输入时须控制格式(即中间用空格隔开)
        fscanf(fp, "%d %s %d", &stu[i].id, &stu[i].name, &stu[i].score);
    }
    if(c==1)
        sort(stu, stu+n, cmp1);
    else if(c==2)
        sort(stu, stu+n, cmp2);
    else    //(c==3)
        sort(stu, stu+n, cmp3);
    for(int i=0; i<n; i++){
        //注意(2): 格式输出 %06d : int占6位,高位补0
        printf("%06d %s %d\n", stu[i].id, stu[i].name, stu[i].score);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/claremore/p/6549557.html

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