Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem.

The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predators, the geese population was out of control. The people of Loowater mostly kept clear of the geese. Occasionally, a goose would attack one of the people, and perhaps bite off a finger or two, but in general, the people tolerated the geese as a minor nuisance.

One day, a freak mutation occurred, and one of the geese spawned a multi-headed fire-breathing dragon. When the dragon grew up, he threatened to burn the Kingdom of Loowater to a crisp. Loowater had a major problem. The king was alarmed, and called on his knights to slay the dragon and save the kingdom.

The knights explained: "To slay the dragon, we must chop off all its heads. Each knight can chop off one of the dragon's heads. The heads of the dragon are of different sizes. In order to chop off a head, a knight must be at least as tall as the diameter of the head. The knights' union demands that for chopping off a head, a knight must be paid a wage equal to one gold coin for each centimetre of the knight's height."

Would there be enough knights to defeat the dragon? The king called on his advisors to help him decide how many and which knights to hire. After having lost a lot of money building Mir Park, the king wanted to minimize the expense of slaying the dragon. As one of the advisors, your job was to help the king. You took it very seriously: if you failed, you and the whole kingdom would be burnt to a crisp!

Input

The input contains several test cases. The first line of each test case contains two integers between 1 and 20000 inclusive, indicating the number n of heads that the dragon has, and the number m of knights in the kingdom. The next n lines each contain an integer, and give the diameters of the dragon's heads, in centimetres. The following m lines each contain an integer, and specify the heights of the knights of Loowater, also in centimetres.

The last test case is followed by a line containing:

0 0

Output

For each test case, output a line containing the minimum number of gold coins that the king needs to pay to slay the dragon. If it is not possible for the knights of Loowater to slay the dragon, output the line:

Loowater is doomed!

Sample Input

2 3
5
4
7
8
4
2 1
5
5
10
0 0

Sample Output

11
Loowater is doomed! 题解: 贪心,每次用最没用代价最小的骑士去打龙,能搞定就搞定,不能搞定就不用他了。 代码:
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#define MAXN 30000
using namespace std;
int n,m;
int d[MAXN],p[MAXN];
int main(){
while(){
memset(d,,sizeof(d));
memset(p,,sizeof(p));
scanf("%d%d",&n,&m);
if(!n&&!m) break;
for(int i=;i<=n;i++) scanf("%d",&d[i]);
for(int i=;i<=m;i++) scanf("%d",&p[i]);
sort(d+,d+n+),sort(p+,p+m+);
int i=,j=,tot=;
while(i<=m&&j<=n){
if(p[i]>=d[j]) j++,tot+=p[i];
i++;
}
if(j<=n) printf("Loowater is doomed!\n");
else printf("%d\n",tot);
}
}

POJ - 3646 The Dragon of Loowater的更多相关文章

  1. The Dragon of Loowater

      The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into ...

  2. uva-----11292 The Dragon of Loowater

    Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance tur ...

  3. uva 11292 Dragon of Loowater (勇者斗恶龙)

    Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance tur ...

  4. UVA它11292 - Dragon of Loowater

    Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance tur ...

  5. UVA 11292 Dragon of Loowater(简单贪心)

    Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance tur ...

  6. Dragon of Loowater

    option=com_onlinejudge&Itemid=8&page=show_problem&problem=2267" style="color:b ...

  7. 贪心/思维题 UVA 11292 The Dragon of Loowater

    题目传送门 /* 题意:n个头,m个士兵,问能否砍掉n个头 贪心/思维题:两个数组升序排序,用最弱的士兵砍掉当前的头 */ #include <cstdio> #include <c ...

  8. Uva11292--------------(The Dragon of Loowater)勇者斗恶龙 (排序后贪心)

    ---恢复内容开始--- 题目: Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major ...

  9. [ACM_水题] UVA 11292 Dragon of Loowater [勇士斗恶龙 双数组排序 贪心]

    Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shor ...

随机推荐

  1. Go依赖管理及Go module使用

    Go语言的依赖管理随着版本的更迭正逐渐完善起来. 依赖管理 为什么需要依赖管理 最早的时候,Go所依赖的所有的第三方库都放在GOPATH这个目录下面.这就导致了同一个库只能保存一个版本的代码.如果不同 ...

  2. CommonsMultipartFile 转为 File 类型

    1.我们可以查看CommonsMultipartFile的源码发现有这样一个方法 @Override public InputStream getInputStream() throws IOExce ...

  3. Linux 笔记 - 第十九章 配置 Squid 正向代理和反向代理服务

    一.简介 Squid 是一个高性能的代理缓存服务器,对应中文的乌贼,鱿鱼的意思.Squid 支持 FTP,gopher 和 HTTP 协议.和一般的代理缓存软件不同,Squid 用一个单独的,非模块化 ...

  4. Cabloy-CMS:动静结合,解决Hexo痛点问题(进阶篇)

    前言 前一篇文章 介绍了如何通过Cabloy-CMS快速搭建一个博客站点. 这里简单介绍Cabloy-CMS静态站点的渲染机制,更多详细的内容请参见https://cms.cabloy.com 渲染规 ...

  5. linux 操作系统级别监控 nmon命令

    nmon是IBM公司开发的Linux性能监控工具,可以实时展示系统性能情况,也可以将监控数据写入文件中,并使用nmon分析器做数据展示 实时监控 命令 ./nmon c 代表CPU m 代表Memor ...

  6. Codeforces Numbers 题解

    这题只需要会10转P进制就行了. PS:答案需要约分,可以直接用c++自带函数__gcd(x,y). 洛谷网址 Codeforces网址 Code(C++): #include<bits/std ...

  7. 暑期——第三周总结(Ubuntu系统安装eclipse问题【已解决】)

    所花时间:7天 代码行:200(python)+150(java) 博客量:1篇 了解到知识点 : 一: Python: 问题 unresolved reference xrange 解决方案 pyt ...

  8. C++ new和malloc的区别

    1.new关键字是C++中的一部分,malloc是由C库提供的函数: 2.new是以具体类型为单位进行内存分配,malloc只能以字节为单位进行内存分配: 3.new在申请单个类型变量时可进行初始化, ...

  9. supervisor配置kibana

    在/etc/supervisor/conf.d/目录下添加kibana.conf [program:kibana]command=/opt/kibana-6.8.1-linux-x86_64/bin/ ...

  10. 检查图片是否损坏、图片后缀是否与实际图片类型对应 - Python

    图片工具 检查图片是否损坏 日常工作中,时常会需要用到图片,有时候图片在下载.解压过程中会损坏,而如果一张一张点击来检查就太不Cool了,因此我想大家都需要一个检查脚本: 测试图片,0.jpg是正常的 ...