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Codeforces Round #233 (Div. 2) A 水题模拟

2014年04月10日 ⁄ 综合 ⁄ 共 1655字 ⁄ 字号 评论关闭
A. Pages
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

User ainta is making a web site. This time he is going to make a navigation of the pages. In his site, there are n pages numbered by integers from 1 to n.
Assume that somebody is on the p-th page now. The navigation will look like this:

<< p - k p - k + 1 ... p - 1 (p) p + 1 ... p + k - 1 p + k >>

When someone clicks the button "<<" he is redirected to page 1,
and when someone clicks the button ">>" he is redirected to page n.
Of course if someone clicks on a number, he is redirected to the corresponding page.

There are some conditions in the navigation:

  • If page 1 is in the navigation, the button "<<" must not be printed.
  • If page n is in the navigation, the button ">>" must not
    be printed.
  • If the page number is smaller than 1 or greater than n, it
    must not be printed.

 

You can see some examples of the navigations. Make a program that prints the navigation.

Input

The first and the only line contains three integers npk (3 ≤ n ≤ 1001 ≤ p ≤ n1 ≤ k ≤ n)

Output

Print the proper navigation. Follow the format of the output from the test samples.

Sample test(s)
input
17 5 2
output
<< 3 4 (5) 6 7 >> 
input
6 5 2
output
<< 3 4 (5) 6 
input
6 1 2
output
(1) 2 3 >> 
input
6 2 2
output
1 (2) 3 4 >>
input
9 6 3
output
<< 3 4 5 (6) 7 8 9
input
10 6 3
output
<< 3 4 5 (6) 7 8 9 >>
input
8 5 4
output
1 2 3 4 (5) 6 7 8 
这么简单的题目还需要解释那么多么?
直接上代码。
/****
	*@PoloShen
	*Title:A
	*/
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;

int n, p, k;

void solve(){
	bool first = 1;
	if (p - k > 1) { printf("<<"); first = 0;}
	for (int i = 1; i <= n; i++){
        if (i == p)
        {
            if (first){ first = 0; printf("(%d)", p); }
            else printf(" (%d)", p);
        }
        else if (p - k <= i && i <= p + k){
            if (first){ first = 0; printf("%d", i); }
            else printf(" %d", i);
        }
        else continue;
	}
	if (p + k < n) printf(" >>\n");
	else printf("\n");
}

int main(){
    while (cin >> n >> p >> k) solve();
    return 0;
}

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