After a long night of coding, Charles Pearson Peterson is having trouble sleeping. This is not only because he is still thinking about the problem he is working on but also due to drinking too much java during the wee hours. This happens frequently, so Charles has developed a routine to count sheep. Not the animal, but the word. Specifically, he thinks of a list of words, many of which are close in spelling to "sheep", and then counts how many actually are the word "sheep". Charles is always careful to be case-sensitive in his matching, so "Sheep" is not a match. You are to write a program that helps Charles count "sheep".

Input

Input will consist of multiple problem instances. The first line will consist of a single positive integer n ≤ 20, which is the number of problem instances. The input for each problem instance will be on two lines. The first line will consist of a positive integer m ≤ 10 and the second line will consist of m words, separated by a single space and each containing no more than 10 characters.

Output

For each problem instance, you are to produce one line of output in the format:

Case i: This list contains n sheep.

The value of i is the number of the problem instance (we assume we start numbering at 1) and n is the number of times the word "sheep" appears in the list of words for that problem instance. Two successive lines should be separated by a single blank line, but do not output any trailing blank line.

Sample Input

4
5
shep sheeps sheep ship Sheep
7
sheep sheep SHEEP sheep shepe shemp seep
10
sheep sheep sheep sheep sheep sheep sheep sheep sheep sheep
4
shape buffalo ram goat

Sample Output

Case 1: This list contains 1 sheep.

Case 2: This list contains 3 sheep.

Case 3: This list contains 10 sheep.

Case 4: This list contains 0 sheep.

Source: East
Central North America 2000 Practice

 #include <stdio.h>
#include <string.h> char str[];
char sheep[] = "sheep"; int main ()
{
int m,n;
int numCount;
scanf("%d",&m);
int i=;
while(i<m)
{
numCount=;
scanf("%d",&n);
getchar();
gets(str);
char *ptr = str;
while(n--)
{
int j;
if((memcmp(ptr,sheep,) == )&&((*(ptr+) == ' ')||(*(ptr+) == '\0')))
{
numCount++;
ptr+=;
}
else
{
for(j=;j<;j++)
{
if(*ptr++ != ' ')
continue;
break;
}
}
}
printf("Case %d: This list contains %d sheep.\n",++i,numCount);
if(i<m)
printf("\n");
numCount = ;
}
return ;
}

2001. Counting Sheep的更多相关文章

  1. 【DFS深搜初步】HDOJ-2952 Counting Sheep、NYOJ-27 水池数目

    [题目链接:HDOJ-2952] Counting Sheep Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 ...

  2. Counting sheep...

    Counting sheep... Description: Consider an array of sheep where some sheep may be missing from their ...

  3. HDU-2952 Counting Sheep (DFS)

    Counting Sheep Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Tota ...

  4. HDU 2952 Counting Sheep(DFS)

    题目链接 Problem Description A while ago I had trouble sleeping. I used to lie awake, staring at the cei ...

  5. HDU2952:Counting Sheep(DFS)

    Counting Sheep Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Tota ...

  6. Hdu2952 Counting Sheep 2017-01-18 14:56 44人阅读 评论(0) 收藏

    Counting Sheep Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Tota ...

  7. ACM HDU-2952 Counting Sheep

    Counting Sheep Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) T ...

  8. 【Kata Daily 190927】Counting sheep...(数绵羊)

    题目: Consider an array of sheep where some sheep may be missing from their place. We need a function ...

  9. hdu 2952 Counting Sheep

    本题来自:http://acm.hdu.edu.cn/showproblem.php?pid=2952 题意:上下左右4个方向为一群.搜索有几群羊 #include <stdio.h> # ...

随机推荐

  1. 【C#】 用Route进行URL重写

    在.NET Framework 4中,微软推出了Route机制.这种机制不仅在MVC中大量运用,在WebForm中也可以使用. 和Contex.RewritePath()一样,Route功能也是写在G ...

  2. .net学习之CTS、CLS和CLR

    CLR:公共语言运行时,就是所有.net语言写的程序的公共运行时环境,比如C#.VB.Net等语言写的程序需要运行在CLR上,然后CLR解析执行操作系统的相关指令,CLR是.net程序运行在操作系统的 ...

  3. Android OkHttp完全解析 --zz

    参考文章 https://github.com/square/okhttp http://square.github.io/okhttp/ 泡网OkHttp使用教程 Android OkHttp完全解 ...

  4. .Net Ioc Unity

    Unity 的接口IUnityContainer public interface IUnityContainer : IDisposable IUnityContainer RegisterType ...

  5. <转>WCF中出现死锁或者超时

    WCF回调中的死锁 一.服务器端死锁 对于如下服务: [ServiceContract(CallbackContract = typeof(INotify))] public class Downlo ...

  6. android 入门-使用adb安装及卸载apk

     我想用adb 安装apk 到设备上现在出现了2个. 提示我没有找到设备    安装不用进去adb shell 这是你存放apk文件夹路径 下面安装apk到手机上(usb一定要连接成功否则读取不到手机 ...

  7. 利用opencv进行相机标定程序

    #include "Stafx.h" ; //棋盘上有13个格子,那么角点的数目12 ; ; //图片的总张数 int main(int argc, char** argv) { ...

  8. URAL 1966 Cycling Roads 点在线段上、线段是否相交、并查集

    F - Cycling Roads     Description When Vova was in Shenzhen, he rented a bike and spent most of the ...

  9. 智能车学习(十八)——电机学习

    一.C车电机选择 1.摘要:      因为C车模在四轮车的优势是有两个电机,可以进行主动差速,劣势是电机太弱了....所以如何选择电机,就是个钱的问题了,电机多一点,就比较好选,但是C车电机跑多了就 ...

  10. sql2014 新建用户并登陆

    EXEC master.dbo.sp_addlogin @loginame = N'testuser1', @passwd = '123456', @defdb = N'master', @defla ...