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. 与你相遇好幸运,Waterline的属性

    >支持的数据类型: string / text / integer / float / date / time / datetime / boolean / binary / array / j ...

  2. LeetCode之Binary Tree Level Order Traversal 层序遍历二叉树

    Binary Tree Level Order Traversal 题目描述: Given a binary tree, return the level order traversal of its ...

  3. .net学习之泛型、程序集和反射

    一.泛型1.CLR编译时,编译器只为MyList<T>类型产生“泛型版”的IL代码——并不进行泛型的实例化,T在中间只充当占位符.例如:MyList 类型元数据中显示的<T> ...

  4. Newtonsoft.Json(Json.Net)学习笔记-高级使用(转)

    1.忽略某些属性 2.默认值的处理 3.空值的处理 4.支持非公共成员 5.日期处理 6.自定义序列化的字段名称 7.动态决定属性是否序列化 8.枚举值的自定义格式化问题 9.自定义类型转换 10.全 ...

  5. 解决MYSQL错误:ERROR 1040 (08004): Too many connections

    方法一: show processlist; show variables like 'max_connections'; show global status like 'max_used_conn ...

  6. 攻城狮在路上(叁)Linux(十七)--- linux磁盘与文件管理概述

    一.复习知识点: 1.扇区是最小的物理存储单位,大小为512bytes. 2.扇区组成一个圆,成为柱面,柱面是分区的最小单位. 3.第一个扇区很重要,因为包含了MBR(446字节)和分区表(64字节) ...

  7. 轻松学习RSA加密算法原理

    转自:http://blog.csdn.net/sunmenggmail/article/details/11994013 http://blog.csdn.net/q376420785/articl ...

  8. java thread run and start

    在java中继承Thread,线程启动有两中方法:start()和run().下面简单介绍一下两者的区别. start():启动一个线程,此时线程处于就绪状态,然后调用Thread对象的run()方法 ...

  9. whl文件安装

    进入whl文件的目录,直接pip install ...即可

  10. Comet:基于 HTTP 长连接的“服务器推”技术解析

    原文链接:http://www.cnblogs.com/deepleo/p/Comet.html 一.背景介绍 传统web请求,是显式的向服务器发送http Request,拿到Response后显示 ...