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# virtual override 和 new 的区别

    一直以来我都对 virtual  override 和 new 之间的区别感到疑惑不解. 特别笔试的时候特别容易考到,真的很容易弄错啊,畜生! 光看理论永远记不住,那不如写几行代码就懂了. 首先看看v ...

  2. RecyclerView导入依赖包

    1. eclipse 上的导入: 如下进入Android SDK的如下路径, \android-sdk\extras\android\m2repository\com\android\support\ ...

  3. ExcelReport第二篇:ExcelReport源码解析

    导航 目   录:基于NPOI的报表引擎——ExcelReport 上一篇:使用ExcelReport导出Excel 下一篇:扩展元素格式化器 概述 针对上一篇随笔收到的反馈,在展开对ExcelRep ...

  4. Web框架之Tornado

    概述 Tornado 是 FriendFeed 使用的可扩展的非阻塞式 web 服务器及其相关工具的开源版本.这个 Web 框架看起来有些像web.py 或者 Google 的 webapp,不过为了 ...

  5. 知乎大牛的关于JS解答

    很多疑惑一扫而空.... http://www.zhihu.com/question/35905242?sort=created JS的单线程,浏览器的多进程,与CPU,OS的对位. 互联网移动的起起 ...

  6. poj 3278:Catch That Cow(简单一维广搜)

    Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 45648   Accepted: 14310 ...

  7. 求CRC校验和的低位和高位的两种方式

    方式1 unsigned ; // 校验和 ]; memcpy(tstCRCChecksum,&shrCRCCheckSum,); // shrCRCCheckSum:216D LOGI(]) ...

  8. visio调整画布大小和旋转画布(转)

    1.调整画布大小: 鼠标移至画布边界(注意不能是顶点附近),按下ctrl,就会出现双向箭头,拖动鼠标即能调整画布大小. 2.旋转画布: 鼠标移至画布顶点或附近,按下ctrl, 出现单箭头优弧,移动鼠标 ...

  9. 【MySQL 安装过程1】顺利安装MySQL完整过程

    一.MySQL Sever的安装 1.开始安装: 2.这里就要开始注意,端口号我们的my SQL端口号为3306 3.下面要输入用户名和用户密码.注意,帐号密码  都是 root. 4.下面的最后一页 ...

  10. MySQL主从复制(Master-Slave)与读写分离(MySQL-Proxy)实践

    主服务器上(注:应该是允许从机访问)  GRANT REPLICATION SLAVE ON *.* to ‘rep1’@’192.168.10.131’ identified by ‘passwor ...