NBUT 1217 Dinner
[1217] Dinner
- 时间限制: 1000 ms 内存限制: 32768 K
- 问题描写叙述
- Little A is one member of ACM team. He had just won the gold in World Final. To celebrate, he decided to invite all to have one meal. As bowl, knife
and other tableware is not enough in the kitchen, Little A goes to take backup tableware in warehouse. There are many boxes in warehouse, one box contains only one thing, and each box is marked by the name of things inside it. For example, if "basketball"
is written on the box, which means the box contains only basketball. With these marks, Little A wants to find out the tableware easily. So, the problem for you is to help him, find out all the tableware from all boxes in the warehouse. - 输入
- There are many test cases. Each case contains one line, and one integer N at the first, N indicates that there are N boxes in the warehouse. Then N strings follow, each string is one name written on the box.
- 输出
- For each test of the input, output all the name of tableware.
- 例子输入
3 basketball fork chopsticks
2 bowl letter- 例子输出
fork chopsticks
bowl- 提示
The tableware only contains: bowl, knife, fork and chopsticks.
- 来源
辽宁省赛2010
题目意思非常easy,就是输入n个字符串。当遇到bowl, knife, fork ,chopsticks.这四个字符串时就输出。所有在一行输出。每两个中间用一个空格隔开,最后不能有空格(我原以为没关系,所以在这里PE了一次…这就是教训啊…)。。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
const int maxn= 100000 + 10;
char str[maxn];
char ss[4][20]= {"bowl","knife","fork","chopsticks"};
int main() {
int n;
while(~scanf("%d",&n)) {
bool flag=0;
for(int i=0; i<n; i++) {
scanf("%s",str);
for(int j=0; j<4; j++)
if(strcmp(str,ss[j])==0) {
if(flag)
printf(" %s",ss[j]);
else
{
printf("%s",ss[j]);flag=1;
}
break;
}
}
puts("");
}
return 0;
}
NBUT 1217 Dinner的更多相关文章
- NBUT 1217 Dinner 2010辽宁省赛
Time limit 1000 ms Memory limit 32768 kB Little A is one member of ACM team. He had just won the g ...
- NBUT 1457 莫队算法 离散化
Sona Time Limit:5000MS Memory Limit:65535KB 64bit IO Format: Submit Status Practice NBUT 145 ...
- nyoj 218 Dinner(贪心专题)
Dinner 时间限制:100 ms | 内存限制:65535 KB 难度:1 描述 Little A is one member of ACM team. He had just won t ...
- ACM: NBUT 1107 盒子游戏 - 简单博弈
NBUT 1107 盒子游戏 Time Limit:1000MS Memory Limit:65535KB 64bit IO Format: Practice Appoint ...
- ACM: NBUT 1105 多连块拼图 - 水题 - 模拟
NBUT 1105 多连块拼图 Time Limit:1000MS Memory Limit:65535KB 64bit IO Format: Practice Appoint ...
- 借教室(codevs 1217)
1217 借教室 2012年NOIP全国联赛提高组 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 钻石 Diamond 题解 查看运行结果 题目描述 Descrip ...
- ACM: NBUT 1646 Internet of Lights and Switches - 二进制+map+vector
NBUT 1646 Internet of Lights and Switches Time Limit:5000MS Memory Limit:65535KB 64bit IO Fo ...
- ural 1217. Unlucky Tickets
1217. Unlucky Tickets Time limit: 1.0 secondMemory limit: 64 MB Strange people live in Moscow! Each ...
- 【wikioi】1217 借教室
题目链接http://www.wikioi.com/problem/1217/ 算法:二分答案(线段树可过wikioi数据) 二分:http://www.wikioi.com/solution/lis ...
随机推荐
- Linux 新建文件/文件夹,删除文件文件夹,查找文件 打开文件
1.新建文件夹:mkdir xx 2.新建文件: touch 1.py 3.删除文件/文件夹: rm -r xx rm 1.py 4.打开文件:cat 1.py 只显示前几行 :head -2 1. ...
- Velocity(5)——#macro 指令
1 #macro(formatIncreaseData $increase) 2 #if(${product.onlineStatusFlag} =='0') 3 -- 4 #elseif(!$inc ...
- [转载] 快速理解Kafka分布式消息队列框架
转载自http://blog.csdn.net/xiaolang85/article/details/18048631 ==是什么 == 简单的说,Kafka是由Linkedin开发的一个分布式的消息 ...
- [转载] java的动态代理机制详解
转载自http://www.cnblogs.com/xiaoluo501395377/p/3383130.html 代理模式 代理模式是常用的java设计模式,他的特征是代理类与委托类有同样的接口,代 ...
- CSharpGL(47)你好,Framebuffer!
CSharpGL(47)你好,Framebuffer! Framebuffer对象(FBO)是一种复杂的OpenGL对象.使用自定义的framebuffer,可以实现离屏渲染,进而实现很多高级功能,例 ...
- phpcms实现全站搜索
如果制作的静态页面中有搜索功能,那么使用phpcms进行替换怎么替换呢?会不会越到很多的麻烦呢?接下来进行phpcms替换静态页面中的搜索功能. 第一步:搜索页面的form表单提交书写,form表单怎 ...
- java-基础-泛型
java泛型通配符问题. java中的泛型基本用法参考<java编程思想>第四版 p.353 java泛型中比较难理解的主要是类型擦除和通配符相关. 1.类型擦除 在编译期间,类型 ...
- 微信小程序开发《三》:微信小程序请求不能使用session的原因及解决办法
本人在前面的微信小程序开发<二>中提到要想在服务端保持状态需要在客户端第一次请求服务器的时候给客户端返回一个sessionid,由客户端在本地保存,下次请求的时候在header里面带上这个 ...
- hadoop2.5的伪分布式安装配置
一.windows环境下安装 根据博主写的一次性安装成功了: http://blog.csdn.net/antgan/article/details/52067441 二.linux环境下(cento ...
- Struts1.2,struts2.0原理分析
struts1原理: 1.首先我们表单提交到action 2.进入到web.xml 3.web.xml拦截*.do 4.交给ActionServlet 5.找到path属性,获得url 6.找到nam ...