1707004731 石文昭

```c
#include
using namespace std;
const int maxn = 10000 + 10;
struct Node
{
int department;//院号
int number;//借书证卡号
char name[maxn];//姓名
char student_id[maxn];//学号
};
Node a[maxn];
void quicksort(Node a[], int left, int right)//按照系名进行排名
{
int i, j;
if (left >= right)
return;

i = left, j = right;
a[0] = a[i];
while (i != j)
{
while (a[j].number > a[0].number && i < j)
--j;
if (i < j)
{
a[i] = a[j];
i++;
}
while (a[i].number < a[0].number && i < j)
++i;
if (i < j)
{
a[j] = a[i];
j--;
}
}
a[i] = a[0];
quicksort(a, left, i - 1);
quicksort(a, i + 1, right);

}

void bin_search(Node a[], int n, int x)
{
int left = 1, right = n;
int ans = 0;
while (left <= right)
{
int mid = (left + right) / 2;
if (a[mid].department == x)
{
ans = mid;
break;
}
else if (a[mid].department < x)
{
left = mid + 1;
}
else
{
right = mid - 1;
}
}
if (!ans)
{
printf("没找到\n");
}
else
{
int start = ans, ends = ans;
while (a[start].department == x)
start--;
while (a[ends].department == x)
ends++;
printf("%d院的信息为：\n", x);
for (int i = start + 1; i < ends; i++)
{
printf("%d %d %s %s\n", a[i].department, a[i].number, a[i].name, a[i].student_id);
}
}
}
int main()
{
int n;
int i;
printf("请输入信息系统学生的个数：\n");
scanf("%d", &n);
printf("请输入学生学院号、卡号、姓名、学号：\n");
for (i = 1; i <= n; i++)
{
scanf("%d%d%s%s", &(a[i].department), &a[i].number, &a[i].name, &a[i].student_id);
}
quicksort(a, 1, n);
printf("按照卡号顺序输出学生信息：\n");
for (i = 1; i <= n; i++)
{
printf("%d %d %s %s\n", a[i].department, a[i].number, a[i].name, a[i].student_id);
}
int x;
printf("请输入要查找的学院号：\n");
scanf("%d", &x);
bin_search(a, n, x);
return 0;
}
```

The issue contains a C program for sorting and searching student records. The problem is not described; the user posted code and an image link. To understand the issue, examine the linked image and the repository context. The code sorts by 'number' but the comment says '按照系名进行排名' (sort by department name). The binary search uses 'department' field. Check for logic errors in sorting and searching, and verify the image for expected output.

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