PAT日志 1028

最新推荐文章于 2021-02-19 14:33:26 发布

doublefloatint

最新推荐文章于 2021-02-19 14:33:26 发布

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分类专栏： PAT日志文章标签： PAT 1028

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本文介绍了一种基于学生记录的排序算法，包括按ID、姓名和成绩进行非减排序的方法。通过定义不同的比较函数，实现了对学生的ID、姓名和成绩进行灵活排序的功能。

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顽强的小白

1028 List Sorting （25 分）

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

题目解析

排序题，三种排序方式

按照ID从小到大，没有重复的ID
按照姓名非减排序，有重复的就按照ID从小到大排
按照分数非减排序，有相同的就按照ID从小到大排

代码实现

struct student{
 int id;
 char name[20];
 int score;
}stu[maxn]; bool cmp1(student a,student b){
 return a.id<b.id;
}
bool cmp2(student a,student b){
 if(strcmp(a.name ,b.name)==0){
  return a.id<b.id;
 }else 
  return strcmp(a.name ,b.name)<0;
}
bool cmp3(student a,student b){
 if(a.score==b.score){
  return a.id<b.id;
 }else 
  return a.score<b.score;
}
int main(){
 int n,tag;
 scanf("%d %d",&n,&tag);
 for(int i=0;i<n;++i){
  scanf("%d %s %d",&stu[i].id,stu[i].name,&stu[i].score);
 }
 if(tag==1){
  sort(stu,stu+n,cmp1);
 }else if (tag==2){
  sort(stu,stu+n,cmp2);
 }else {
  sort(stu,stu+n,cmp3);
 }
 for(int i=0;i<n;++i){
  printf("%06d %s %d\n",stu[i].id,stu[i].name,stu[i].score);
 }
 return 0;
}