KIDx's Pagination

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本文介绍了一个简单的分页器模拟题目，通过给定的总页数、当前页及显示范围来生成页面导航栏。代码示例使用 C 语言实现，展示了如何根据规则动态生成分页按钮。

KIDx's Pagination

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Submit Statistic Next Problem

Problem Description

One Day, KIDx developed a beautiful pagination for ACdream. Now, KIDx wants you to make another one.

The are n pages in total.
The current page is cur.
The max distance to current page you can display is d.

Here are some rules:

The cur page button is disabled.
If cur page is the first page, the button "<<" should be disabled.
If cur page is the last page, the button ">>" should be disabled.
If the button "x" is disabled, print "[x]"; else print "(x)".
You should not display the "..." button when there is no hidden page.

You can assume that the button "..." is always disabled.

Input

There are multiple cases.
Ease case contains three integers n, cur, d.
1 ≤ n ≤ 100.
1 ≤ cur ≤ n.
0 ≤ d ≤ n.

Output

For each test case, output one line containing "Case #x: " followed by the result of pagination.

Sample Input

10 5 2
10 1 2

Sample Output

Case #1: (<<)[...](3)(4)[5](6)(7)[...](>>)
Case #2: [<<][1](2)(3)[...](>>)

Hint

Case 1:

Case 2:

Source

KIDx

Manager

KIDx

///这是一道简单的模拟题，可我>_<||

#include<stdio.h>
#include<string.h>
#include<math.h>

int main()
{
    int n, cur, d, num=0;
    while(~scanf("%d%d%d", &n, &cur, &d))
    {
        num++;
        printf("Case #%d: ", num);
        if(cur==1)
            printf("[<<]");
        else
            printf("(<<)");
        if(cur-d>1)
            printf("[...]");
        for(int i=1;i<=n;i++)
        {
            if(i==cur)
                printf("[%d]", cur);
            else if(fabs(cur-i)<=d)
                printf("(%d)", i);
        }
        if(cur+d<n)
            printf("[...]");
        if(cur==n)
            printf("[>>]");
        else
            printf("(>>)");
        puts("");
    }
    return 0;