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 recor1028?List Sortingds 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

解題思路：

理解題意：給出一組數據，根據第一行給出的兩個數，第一個數表示有多少行數據，第二個數是按多少列排序

本段代碼分為3段：首先，根據題意輸入數據，其次，根據題意用不同的排序函數排序；最后輸出數據

總結：本題十分簡單，理解即可，只要根據不同的要求做不同的排序即可

代碼：

#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; struct stu{int id;char name[9];int grade; }s[100010];bool cmp1(stu a,stu b){ return a.id<b.id; } bool cmp2(stu a,stu b){int s= strcmp(a.name,b.name);if(s!=0) return s<0;else return a.id<b.id; }bool cmp3(stu a,stu b){if(a.grade!=b.grade) return a.grade<b.grade;else return a.id<b.id; } int main(){int N,C;scanf("%d%d",&N,&C);for(int i=0;i<N;i++){scanf("%d",&s[i].id);scanf("%s %d",s[i].name,&s[i].grade);}if(C==1) sort(s,s+N,cmp1);else if(C==2) sort(s,s+N,cmp2);else sort(s,s+N,cmp3);for(int i=0;i<N;i++){printf("%06d %s %d\n",s[i].id,s[i].name,s[i].grade);}return 0; }

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