1028. List Sorting (25)
题目链接:http://www.patest.cn/contests/pat-a-practise/1028
题目:
1028. List Sorting (25)
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 13 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 90Sample Output 3
000001 Zoe 60 000007 James 85 000002 James 90 000010 Amy 90
分析:
接收数字来确定排序的方案,写好结构体和排序就OK,难度不大
AC代码:
#include<stdio.h> #include<vector> #include<string.h> #include<algorithm> using namespace std; struct Student{ char ID[7]; char name[9]; int grade; }; bool cmp1(Student A, Student B){//依照学号排序 return strcmp(A.ID, B.ID) < 0; } bool cmp2(Student A, Student B){//依照名字排序。再依照学号排序 if (strcmp(A.name, B.name) != 0)return strcmp(A.name, B.name) < 0; else return strcmp(A.ID, B.ID) < 0; } bool cmp3(Student A, Student B){//依照分数排序,再依照学号排序 if (A.grade != B.grade)return A.grade < B.grade; else return strcmp(A.ID, B.ID) < 0; } vector<Student>V; int main(void){ //freopen("F://Temp/input.txt", "r", stdin); int n, k; while (scanf("%d", &n) != EOF){ scanf("%d", &k); for (int i = 0; i < n; i++){ Student tmp; scanf("%s%s%d", tmp.ID, tmp.name, &tmp.grade); V.push_back(tmp); } if (k == 1)sort(V.begin(), V.end(), cmp1); if (k == 2)sort(V.begin(), V.end(), cmp2); if (k == 3)sort(V.begin(), V.end(), cmp3); for (int i = 0; i < n; i++){ printf("%s %s %d\n", V[i].ID, V[i].name, V[i].grade); } } return 0; }
截图:
——Apie陈小旭