AOJ-AHU-OJ-31 Counting Sheep

最新推荐文章于 2022-10-29 19:04:43 发布

J_Sure

最新推荐文章于 2022-10-29 19:04:43 发布

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分类专栏：字符串 ACM-AHU 文章标签： aoj c++ ahu oj c语言

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Counting Sheep

Time Limit: 1000 ms Case Time Limit: 1000 ms Memory Limit: 64 MB

Description

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. Output lines
should be separated by a single blank line.

Sample Input

Original

Transformed

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

Original

Transformed

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 Practise

———————————————————————分割线————————————————————————

思路：极其简单的一个字符串比较函数的应用。

代码如下：

#include <stdio.h>
#include <string.h>
char real[6] = "sheep";
int main(){
	int T, n;
	int i, j = 1;
	scanf("%d", &T);
	while(T--){
		int count = 0;
		scanf("%d", &n);
		char sheep[n][11];
		for(i = 0; i < n; i++)
		  scanf("%s", sheep[i]);
		getchar();
		for( i = 0; i < n; i++){
		  if (strcmp(sheep[i], real) == 0)
		    count++;				
		}
		printf("Case %d: This list contains %d sheep.\n", j++, count);
		if(T != 0)
		putchar('\n');	
	}
	return 0;
}