PAT-A1028 List Sorting 题目内容及题解
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
题目大意
题目给定一组数据,并根据题目要求排序并输出。
解题思路
- 排序输出并返回0值即可。
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 100010
struct Student{
int id,grade;
char name[10];
}stu[maxn];
bool cmp1(Student a,Student b){
return a.id<b.id;
}
bool cmp2(Student a,Student b){
int i=strcmp(a.name,b.name);
if(i!=0){
return i<0;
}else{
return a.id<b.id;
}
}
bool cmp3(Student a,Student b){
if(a.grade!=b.grade){
return a.grade<b.grade;
}else{
return a.id<b.id;
}
}
int main(){
int N,C;
int i;
scanf("%d%d",&N,&C);
for(i=0;i<N;i++){
scanf("%d %s %d",&stu[i].id,&stu[i].name,&stu[i].grade);
}
if(C==1){
sort(stu,stu+N,cmp1);
}else if(C==2){
sort(stu,stu+N,cmp2);
}else{
sort(stu,stu+N,cmp3);
}
for(i=0;i<N;i++){
printf("%06d %s %d\n",stu[i].id,stu[i].name,stu[i].grade);
}
return 0;
}
运行结果