List Sorting（注意输出格式）

题目

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

答案

#include
using namespace std;
struct Node{
     
	int num,grade;
	string name; 
};

bool cmp1(Node x,Node y){
     
	return x.num<y.num;
}

bool cmp2(Node x,Node y){
     
	if(x.name==y.name) return x.num<y.num;
	else return x.name<y.name;
}

bool cmp3(Node x,Node y){
     
	if(x.grade==y.grade) return x.num<y.num;
	else return x.grade<y.grade;
}

int main()
{
     
	int n,c;
	cin>>n>>c;
	Node node[n];
	for(int i=0;i<n;i++)
	cin>>node[i].num>>node[i].name>>node[i].grade;
	if(c==1) sort(node,node+n,cmp1);
	else if(c==2) sort(node,node+n,cmp2);
	else if(c==3) sort(node,node+n,cmp3);
	for(int i=0;i<n;i++)
	{
     
		printf("%06d",node[i].num);
		cout<<" "<<node[i].name<<" "<<node[i].grade<<endl;
	}
}

注意

最终要注意输出6位学号，学号不足6位高位补0