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 9
## Sample Output 3:
```c
000002 James 9
000001 Zoe 60
000007 James 85
000010 Amy 90
思路核心
会写bmp比较sort函数就行了，其余按部就班。
完整源码
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct S{
char id[8];
char name[10];
int grade;
};
int m;
bool cmp(S a,S b){
if(m==2){
if(strcmp(a.name,b.name)!=0) return strcmp(a.name,b.name)<0;
}
if(m==3){
if(a.grade!=b.grade) return a.grade<b.grade;
}
return strcmp(a.id,b.id)<0;
}
int main()
{
int n;
scanf("%d %d",&n,&m);
struct S s[n];
for (int i = 0; i < n; i++) {
scanf("%s %s %d", s[i].id, s[i].name,&s[i].grade);
}
sort(s,s+n,cmp);
for(int i =0;i<n;i++){
printf("%s %s %d\n",s[i].id,s[i].name,s[i].grade);
}
return 0;
}
测试效果
