CF Sea and Islands
1 second
256 megabytes
standard input
standard output
A map of some object is a rectangular field consisting of n rows and n columns. Each cell is initially occupied by the sea but you can cover some some cells of the map with sand so that exactly k islands appear on the map. We will call a set of sand cells to be island if it is possible to get from each of them to each of them by moving only through sand cells and by moving from a cell only to a side-adjacent cell. The cells are called to be side-adjacent if they share a vertical or horizontal side. It is easy to see that islands do not share cells (otherwise they together form a bigger island).
Find a way to cover some cells with sand so that exactly k islands appear on the n × n map, or determine that no such way exists.
The single line contains two positive integers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ n2) — the size of the map and the number of islands you should form.
If the answer doesn't exist, print "NO" (without the quotes) in a single line.
Otherwise, print "YES" in the first line. In the next n lines print the description of the map. Each of the lines of the description must consist only of characters 'S' and 'L', where 'S' is a cell that is occupied by the sea and 'L' is the cell covered with sand. The length of each line of the description must equal n.
If there are multiple answers, you may print any of them.
You should not maximize the sizes of islands.
5 2
YES
SSSSS
LLLLL
SSSSS
LLLLL
SSSSS
5 25
NO 判下奇偶再输出就行了。人品爆发居然打到了200+,希望这样的狗屎运常来一些。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <map>
#include <ctime>
using namespace std; int main(void)
{
int n,k;
int max_land; cin >> n >> k; if(n % )
max_land = (n / + + n / ) * (n / ) + n / + ;
else
max_land = n * n / ;
if(max_land < k)
puts("NO");
else
{
puts("YES");
if(n % == )
{
int flag = ;
int count = ;
for(int i = ;i < n;i ++)
{
for(int j = ;j < n;j ++)
if(count < k && (j + flag) % == )
{
cout << 'L';
count ++;
}
else
cout << 'S';
flag = !flag;
cout << endl;
}
}
else
{
int flag = ;
int count = ;
for(int i = ;i < n;i ++)
{
for(int j = ;j < n;j ++)
{
if(count < k && !flag && j % == )
{
cout << 'L';
count ++;
}
else if(count < k && flag && j % )
{
cout << 'L';
count ++;
}
else
cout << 'S';
}
flag = flag == ? : ;
cout << endl;
} }
} return ;
}
CF Sea and Islands的更多相关文章
- 构造 Codeforces Round #302 (Div. 2) B Sea and Islands
题目传送门 /* 题意:在n^n的海洋里是否有k块陆地 构造算法:按奇偶性来判断,k小于等于所有点数的一半,交叉输出L/S 输出完k个L后,之后全部输出S:) 5 10 的例子可以是这样的: LSLS ...
- Codeforces Round #302 (Div. 2) B. Sea and Islands 构造
B. Sea and Islands Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/544/p ...
- B - Sea and Islands
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Description A map ...
- 544B. Sea and Islands
题目链接 题意: n*n的里面全是S的方格中,填充L,若填充的L上下左右都没有相邻的L则是一个快,问题是能否形成k个块 n可以去奇数也可以去偶数 只要我们输出满足条件的一个结果就好了 对于从0 - n ...
- [Codeforces 505C]Mr. Kitayuta, the Treasure Hunter
Description The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. The is ...
- | dp-the Treasure Hunter
题目: A. Mr. Kitayuta, the Treasure Hunter time limit per test 1 second memory limit per test 256 mega ...
- Codeforces Round #302 解题报告
感觉今天早上虽然没有睡醒但是效率还是挺高的... Pas和C++换着写... 544A. Set of Strings You are given a string q. A sequence o ...
- Codeforces Round #302 (Div. 2)
A. Set of Strings 题意:能否把一个字符串划分为n段,且每段第一个字母都不相同? 思路:判断字符串中出现的字符种数,然后划分即可. #include<iostream> # ...
- codeforces 505C C. Mr. Kitayuta, the Treasure Hunter(dp)
题目链接: C. Mr. Kitayuta, the Treasure Hunter time limit per test 1 second memory limit per test 256 me ...
随机推荐
- F5 负载均衡 相关资源
F5负载均衡之检查命令的说明http://net.zdnet.com.cn/network_security_zone/2010/0505/1730942.shtml F5培训http://wenku ...
- 在Button的click事件中引起客户端JavaScript
void action1_Execute(object sender, SimpleActionExecuteEventArgs e) { WebWindow.CurrentRequestWindow ...
- (剑指Offer)面试题17:合并两个排序的链表
题目: 输入两个递增排序的链表,合并这两个链表并使新链表中的结点仍然时按照递增排序的. 链表结点定义如下: struct ListNode{ int val; ListNode* next; }; 思 ...
- spring 切面 前置后置通知 环绕通知demo
环绕通知: <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http:// ...
- WPF中的ControlTemplate(控件模板)(转)
原文地址 http://www.cnblogs.com/zhouyinhui/archive/2007/03/28/690993.html WPF中的ControlTemplate(控件模板) ...
- 分布式服务框架 Zookeeper -- 管理分布式环境中的数据(转载)
本文转载自:http://www.ibm.com/developerworks/cn/opensource/os-cn-zookeeper/ Zookeeper 分布式服务框架是 Apache Had ...
- 【M9】利用destructors避免泄漏资源
1.在堆上获取的动态资源,用户忘记delete,或者由于异常导致没有没执行到delete,都会造成资源泄漏. 2.我们知道,栈上的对象,离开作用域,必定要执行析构方法.即使抛出异常,会堆栈回滚,保证已 ...
- svn cleanup failed–previous operation has not finished 解决方法
今天svn遇到一个头疼的问题,最开始更新的时候失败了,因为有文件被锁住了.按照以往的操作,我对父目录进行clean up操作,但是clean up 操作也失败了! svn cleanup failed ...
- Android操作系统服务(Context.getSystemService() )
getSystemService是Android很重要的一个API,它是Activity的一个方法,根据传入的NAME来取得对应的Object,然后转换成相应的服务对象.下面介绍系统相应的服务: 传入 ...
- 高级I/O之I/O多路转接——pool、select
当从一个描述符读,然后又写到另一个描述符时,可以在下列形式的循环中使用阻塞I/O: ) if (write(STDOUT_FILENO, buf, n) != n) err_sys("wri ...