PAT A1028

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原创最新推荐文章于 2026-05-21 09:20:05 发布 386 阅读 AI 写同款· GEO 优化 ›

深度伪造检测实战：轻量级频域特征 pipeline 部署指南深度伪造检测（deepfake detection）是多媒体内容安全的核心技术，其本质是通过建模生成式篡改在像素与频域留下的微弱异常痕迹。传统端到端大模型虽在标准数据集上表现优异，却因训练域与真实场景失配、计算开销高、缺乏可解释性而难以落地。本文聚焦于工程可复现的实践路径，提出‘人脸定位→频域伪造特征提取（Xception-FCN + Frequency-Aware Masking）→轻量分类决策’三段式解耦架构，兼顾精度、延迟与可解释性。该方案已在边缘设备（Jetson Orin NX）实现单帧<48ms 阅读详情

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;
}