Kindergarten Counting Game - UVa494

最新推荐文章于 2016-08-17 09:02:13 发布

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最新推荐文章于 2016-08-17 09:02:13 发布

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原文链接：http://www.cnblogs.com/milkcu/p/3808853.html

本文详细介绍了UVA494题目“Kindergarten Counting Game”的解题思路及代码实现，通过状态标志来判断单词边界，从而准确统计每行中的单词数量。

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原文地址：http://www.milkcu.com/blog/archives/uva494.html

题目描述

Kindergarten Counting Game

Everybody sit down in a circle. Ok. Listen to me carefully.

``Woooooo, you scwewy wabbit!''

Now, could someone tell me how many words I just said?

Input and Output

Input to your program will consist of a series of lines, each line containing multiple words (at least one). A ``word'' is defined as a consecutive sequence of letters (upper and/or lower case).

Your program should output a word count for each line of input. Each word count should be printed on a separate line.

Sample Input

Meep Meep!
I tot I taw a putty tat.
I did! I did! I did taw a putty tat.
Shsssssssssh ... I am hunting wabbits. Heh Heh Heh Heh ...

Sample Output

解题思路

这个题目要统计每行中的单词数目。
注意对输入的处理。
使用了一个状态标志state：
state为1表示上一字符在单词内；
state为0表示上一字符不在单词内。

代码实现

#include <stdio.h>
int isl(char c) {
	if((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')) {
		return 1;
	} else {
		return 0;
	}
}
int main(void) {
	char s[100000];
	int rear = 0;
	int c;
	while((c = getchar()) != EOF) {
		s[rear++] = c;
	}
	s[rear] = '\0';
	int i = 0;
	int num = 0;
	int state = 0;
	while(s[i] != '\0') {
		if(s[i] == '\n') {
			printf("%d\n", num);
			num = 0;
			state = 0;
		} else {
			if(isl(s[i])) {
				if(state == 0) {
					num++;
				}
				state = 1;
			} else {
				state = 0;
			}
		}
		i++;
	}
	return 0;
}

（全文完）

转载于:https://www.cnblogs.com/milkcu/p/3808853.html