PAT日志 1028

頑強的小白

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 (≤10​5​​) 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

題目解析

排序題，三種排序方式

• 按照ID從小到大，沒有重複的ID
• 按照姓名非減排序，有重複的就按照ID從小到大排
• 按照分數非減排序，有相同的就按照ID從小到大排

代碼實作

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;
}