PAT A1028

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 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
实现代码如下:

#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
struct Student {
    int id;
    char name[10];
    int score;
}stu[100005];
bool cmp1(struct Student a, struct Student b) {
    return a.id < b.id;
}
bool cmp2(struct Student a, struct Student b) {
    int k = strcmp(a.name, b.name);
    if(k != 0) return k < 0;
    else return a.id < b.id;
}
bool cmp3(struct Student a, struct Student b) {
    if(a.score != b.score) return a.score < b.score;
    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 %d", &stu[i].id, stu[i].name, &stu[i].score);
    }
    if(C == 1) sort(stu, stu + N, cmp1);
    else if(C == 2) sort(stu, stu + N, cmp2);
    else sort(stu, stu + N, cmp3);

    for(int i = 0; i < N; i++){
        printf("%06d %s %d\n", stu[i].id, stu[i].name, stu[i].score);
    }
    return 0;
}





PAT乙级1028题目的名称是“检查密码”,此题目要求考生编写一个程序来验证给定的一组字符串是否符合特定的安全标准。 对于这道题,安全规则如下: - 密码长度需介于6至20之间; - 至少包含一个小写字母、大写字母以及数字; - 不能包括其他字符; - 同一字母连续出现次数不超过两个; 为了帮助理解如何解决这个问题,可以提供一段伪代码逻辑用于处理这类问题: ```plaintext for each password in the list of passwords: if length(password) is not between 6 and 20: mark as invalid continue to next password initialize counters for lowercase, uppercase, digit, other characters, and consecutive count previous_character = None for each character in password: increment appropriate counter based on character type if character equals previous_character: increase consecutive count if consecutive count exceeds two: mark as invalid break from loop else: reset consecutive count set previous_character to current character after checking all characters, if any counter (except 'other') remains at zero or there were too many consecutive characters: mark as invalid otherwise: mark as valid ``` 这段伪代码展示了基本的思路,但实际编程时需要转换成具体的编程语言语法,并考虑边界条件和其他细节以确保正确性。此外,在PAT考试中,效率也是一个考量因素,因此应该尽量优化解决方案减少不必要的计算。 关于具体实现,可以根据所使用的编程语言调整上述逻辑。例如,在Python中可以用正则表达式简化某些判断过程。
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