PAT-A-1028. List Sorting (25)

1028. List Sorting (25)

时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) 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

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

只是一道简单的题目,说实话PAT甲也就30分题目会难一点。

这道题就是注意用cout和cin可能会超时,还是用scanf和printf比较快,其他没有问题了。

#include #include #include #include #include #include #include #include #include using namespace std; struct node{ int id; char name[20]; int grade; }stu[1000100]; bool cmp1(node a,node b){ return a.id if(a.grade==b.grade) return a.id else return a.grade int n,c; cin>>n>>c; for(int i=0;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,com2); else  sort(stu,stu+n,cmp3); for(int i=0;i printf("%06d %s %d\n",stu[i].id,stu[i].name,stu[i].grade); } system("pause"); return 0; }

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