周赛-KIDx's Pagination 分类： 比赛 2015-08-02 08:23 7人阅读 评论(0) 收藏

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: 

周赛的一道模拟

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <algorithm>
#define LL long long
#define RR freopen("output.txt","r",stdin)
#define WW freopen("input.txt","w",stdout)

using namespace std;
bool vis[120];
int main()
{
    int n,cur,d;
    int w=1;
    while(~scanf("%d %d %d",&n,&cur,&d))
    {
        memset(vis,false,sizeof(vis));
        for(int i=1; i<=n; i++)
        {
            if((cur-i<=d&&cur-i>=0)||(i-cur<=d&&i-cur>=0))
            {
                vis[i]=true;
            }
        }
        printf("Case #%d: ",w++);
        if(cur==1)
        {
            printf("[<<]");
        }
        else
        {
            printf("(<<)");
        }
        for(int i=1; i<=n; i++)
        {
            if(i==1)
            {
                if(vis[i])
                {
                    if(cur==i)
                        printf("[%d]",i);
                    else
                    {
                        printf("(%d)",i);
                    }
                }
                else
                {
                    printf("[...]");
                }
            }
            else if(i==n)
            {
                if(vis[n])
                {
                    if(cur==n)
                        printf("[%d]",i);
                    else
                    {
                        printf("(%d)",i);

                    }
                }
                else
                {
                    printf("[...]");
                }
            }
            else if(vis[i])
            {
                if(cur==i)
                printf("[%d]",i);
                else
                {
                    printf("(%d)",i);
                }

            }
        }
        if(cur==n)
        {
            printf("[>>]");
        }
        else
        {
            printf("(>>)");

        }
        printf("\n");
    }


    return 0;
}

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