1028 List Sorting (25 分)
Excel can sort records according to any column. Now you are supposed to imitate this function.
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 Nlines 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).
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.
3 1
000007 James 85
000010 Amy 90
000001 Zoe 60
000001 Zoe 60
000007 James 85
000010 Amy 90
4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98
000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60
4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90
000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90
给出N个学生的ID,姓名以及成绩,按照三种方式排序。
C==1时,按照ID递增的顺序排序,C==2时按照姓名非递减的顺序排序,C==3按照成绩非递减的顺序排序。
若学生姓名和成绩相同,则按照学号递增的顺序排序
写三个comp() 然后用sort()排序。
#include
using namespace std;
struct Student{
int id;
string name;
int score;
}stu[100010];
bool comp1(Student stu1,Student stu2){
return stu1.id < stu2.id;
}
bool comp2(Student stu1,Student stu2){
if(stu1.name.compare(stu2.name)<0) return true;
else if(stu1.name.compare(stu2.name)==0){
if(stu1.id < stu2.id) return true;
}
return false;
}
bool comp3(Student stu1,Student stu2){
if(stu1.score < stu2.score) return true;
else if(stu1.score == stu2.score){
if(stu1.id < stu2.id) return true;
}
return false;
}
int main(){
int C,N,i;
cin>>N>>C;
for(i=0;i>stu[i].id>>stu[i].name>>stu[i].score;
}
switch(C){
case 1:sort(stu,stu+N,comp1);break;
case 2:sort(stu,stu+N,comp2);break;
case 3:sort(stu,stu+N,comp3);break;
}
for(i=0;i
2019.2.8
#include
using namespace std;
struct Rec{
int id,grade;
char name[10];
};
Rec rec[100010];
bool com1(Rec r1,Rec r2){
return r1.id