PAT (Advanced Level) Practice — 1028 List Sorting （25 分）

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本文介绍了一个简单的排序算法，用于根据指定列对学生记录进行排序。包括ID、姓名和成绩的非递减排序，当姓名或成绩相同时，按ID递增排序。

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805468327690240

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

：题意 很简单的题目，如果c==1，按编号的递增排序，如果c==2，按名字的非递减排序，如果c==3，按成绩的非递减排序。

如果名字或者成绩相同，那么按编号的递增排序。

如果用cin cout 最后一个测试点会超时...

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
struct node{
	char no[10];
	char na[10];
	int gra;
}p[111111];
bool cmp1(node x,node y){
	if(strcmp(x.no,y.no)){
		return strcmp(x.no,y.no)<0; 
	}
}
bool cmp2(node x,node y){
	if(strcmp(x.na,y.na)){
		return strcmp(x.na,y.na)<0;
	}else{
		return strcmp(x.no,y.no)<0;
	}
}
bool cmp3(node x,node y){
	if(x.gra!=y.gra){
		return x.gra<y.gra;
	}else{
		return strcmp(x.no,y.no)<0;
	}
}
int main(){
	int n,c;
	scanf("%d%d",&n,&c);
	for(int i=0;i<n;i++){
		//cin>>p[i].no>>p[i].na>>p[i].gra;
		scanf("%s%s%d",p[i].no,p[i].na,&p[i].gra);
	}
	if(c==1){
		sort(p,p+n,cmp1);
	}else if(c==2){
		sort(p,p+n,cmp2);
	}else if(c==3){
		sort(p,p+n,cmp3);
	}
	for(int i=0;i<n;i++){
		//cout<<p[i].no<<" "<<p[i].na<<" "<<p[i].gra<<endl;
		printf("%s %s %d\n",p[i].no,p[i].na,p[i].gra);
	}
	return 0;
}