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 (≤10e5) 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 60Sample Output 1:
000001 Zoe 60
000007 James 85
000010 Amy 90Sample Input 2:
4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98Sample Output 2:
000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60Sample Input 3:
4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 9Sample Output 3:
000002 James 9
000001 Zoe 60
000007 James 85
000010 Amy 90 题目大意：给出n个记录，分别包括id，name，grade。根据c的值是1还是2还是3，对相应的列排序，其中第一列升序，第二列不降序，第三列不降序。
分析：直接用stl的sort排序即可。
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <cmath>
using namespace std;typedef struct record
{int id,grade;string name;
}record;bool cmp1(record a,record b)
{if(a.id!=b.id)return a.id<b.id;
}bool cmp2(record a,record b)
{if(a.name!=b.name)return a.name<=b.name;return a.id<b.id;
}bool cmp3(record a,record b)
{if(a.grade!=b.grade)return a.grade<=b.grade;return a.id<b.id;
}int main(void)
{#ifdef testfreopen("in.txt","r",stdin);//freopen("in.txt","w",stdout);clock_t start=clock();#endif //testint n,c;scanf("%d%d",&n,&c);record re[n+5];for(int i=0;i<n;++i){scanf("%d",&re[i].id);cin>>re[i].name;scanf("%d",&re[i].grade);}switch (c){case 1:sort(re,re+n,cmp1);break;case 2:sort(re,re+n,cmp2);break;case 3:sort(re,re+n,cmp3);break;}for(int i=0;i<n;++i){printf("%06d ",re[i].id);cout<<re[i].name;printf(" %d\n",re[i].grade);}#ifdef testclockid_t end=clock();double endtime=(double)(end-start)/CLOCKS_PER_SEC;printf("\n\n\n\n\n");cout<<"Total time:"<<endtime<<"s"<<endl; //s为单位cout<<"Total time:"<<endtime*1000<<"ms"<<endl; //ms为单位#endif //testreturn 0;
}