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. Elasticsearch在Windows下的安装

    下载Elasticsearch,地址:elasticsearch.org/download 下载jdk,百度搜索jdk下载即可 配置JAVA_HOME变量,配置方法在此文:http://jingyan ...

  2. css2

    CSS 实现div宽度根据内容自适应 <!DOCTYPE html> <html> <head> <meta charset="utf-8" ...

  3. hdu 5288 数学 ****

    给一个序列 定义函数f(l ,r) 为区间[l ,r] 中 的数ai不是在这个区间其他任意数aj的倍数 求所有f(l,r)之和 通过预处理,记录 a[i] 的左右边界(所谓的左右边界时 在从 a[i] ...

  4. [Tools] Eclipse使用小技巧-持续更新

    [背景] 使用之中发现一些eclipse使用的小技巧,记录下来供以后查阅   Eclipse保存preferences,并导入到其他workspaces The Export wizard can b ...

  5. TI Zigbee Light Link 参考设计

    TI  Zigbee Light Link 参考设计 原文出处: http://processors.wiki.ti.com/index.php/Category:ZigBee_Light_Link ...

  6. leetcode4568

    date: 2015-09-13 16:32:49 Median of Two Sorted Arrays There are two sorted arrays nums1 and nums2 of ...

  7. 用C#基于WCF创建TCP的Service供Client端调用

    本文将详细讲解用C#基于WCF创建TCP的Service供Client端调用的详细过程 1):首先创建一个Windows Service的工程 2):生成的代码工程结构如下所示 3):我们将Servi ...

  8. Excel动态合并行、合并列

    背景: 在北京工作的时候,又一次同事问了我这样一个问题,说我要把从数据库获取到的数据直接通过NPOI进行导出,但是我对导出的格式要特殊的要求,如图: 冥思苦想,最终顺利帮同事解决问题,虽然有点瑕疵,但 ...

  9. 【项目经验】 Html Select 遇上 Easyui

    一.背景: 当我在做课表选择触发事件的时候,我发现了一个问题,就是我们直接用的easyui-combobox里面的的绑定事件(onchange)貌似触发不了,这是为什么呢? 二.结论及方法 .原始方法 ...

  10. android常用对话框封装

    在android开发中,经常会用到对话框跟用户进行交互,方便用户可操作性:接下来就对常用对话框进行简单封装,避免在项目中出现冗余代码,加重后期项目的维护量:代码如有问题欢迎大家拍砖指正一起进步. 先贴 ...