1041 考试座位号（附详细注释，逻辑分析）

写在前面

• 实现思路

• 结构体封装学生元信息，试机座位号作为下标
• long long 处理准考证号（16 位数字），也可用string

• 题目较为简单，熟练情况下10分钟a题

测试用例

input:
4
3310120150912233 2 4
3310120150912119 4 1
3310120150912126 1 3
3310120150912002 3 2
2
3 4

output:
3310120150912002 2
3310120150912119 1

ac代码

• C语言版​​long long​​

#include<cstdio>
const int maxn = 1010;
struct Student
{
    long long id;
    int examSeat;
} testSeat[maxn];

int main()
{
    int n, m, seat, examSeat;
    long long id;
    scanf("%d", &n);
    for(int i=0; i<n; i++)
    {
        scanf("%lld%d%d", &id, &seat, &examSeat);
        testSeat[seat].id = id;
        testSeat[seat].examSeat = examSeat;
    }

    scanf("%d", &m);
    for(int i=0; i<m; i++)
    {
        scanf("%d", &seat);
        printf("%lld %d\n", testSeat[seat].id, testSeat[seat].examSeat);
    }

    return 0;
}

• C++版​​string​​···个人尝试···

#include<iostream>
using namespace std;

const int maxn = 1010;
struct Student
{
    string id;
    int examSeat;
} testSeat[maxn];

int main()
{
    int n, m, seat, examSeat;
    char id[16];
    scanf("%d", &n);
    for(int i=0; i<n; i++)
    {
        scanf("%s%d%d", id, &seat, &examSeat);
        testSeat[seat].id = id;
        testSeat[seat].examSeat = examSeat;
    }

    scanf("%d", &m);
    for(int i=0; i<m; i++)
    {
        scanf("%d", &seat);
        printf("%s %d\n", testSeat[seat].id.c_str(), testSeat[seat].examSeat);
    }
    return 0;
}

参考代码

• ​​C++ string 二维数组版···值得参考···​​

#include <iostream>
using namespace std;
int main() {
    string stu[1005][2], s1, s2;;
    int n, m, t;
    cin >> n;
    for(int i = 0; i < n; i++) {
        cin >> s1 >> t >> s2;
        stu[t][0] = s1;
        stu[t][1] = s2;
    }
    cin >> m;
    for(int i = 0; i < m; i++) {
        cin >> t;
        cout << stu[t][0] << " " << stu[t][1] << endl;
    }
    return 0;
}