Codeforces Beta Round #65 (Div. 2) / 71A Way Too Long Words(字符串操作)

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原创最新推荐文章于 2026-02-27 14:07:23 发布 1.2k 阅读 AI 写同款· GEO 优化 ›

codeforces71A-C语言解题报告 71A题目地址题目解答 1.输入单词超过10个字母的单词,输出第一个字母+中间个数+最后一个字母没有超过的,直接输出 2.循环读取输入在for循环内部,接收用户输入的单词知识点 1.接收用户输入 int a; scanf("%d",&a); 2.字符串数组 C语言中没有专门的字符串数组,通过字符数组的方式存放 char arr[100]; 3.计算字符串长度 #include<string.h> int str=strlen(arr); 4.C语言中没有截取字符串的方法,所以使阅读详情

Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.

Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.

This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.

Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".

You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.

Input

The first line contains an integer n (1 ≤ n ≤ 100). Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.

Output

Print n lines. The i-th line should contain the result of replacing of the i-th word from the input data.

   Sample test(s)
 

     input
   
4
word
localization
internationalization
pneumonoultramicroscopicsilicovolcanoconiosis

     output
   
word
l10n
i18n
p43s

注意使用gets前读掉'\n'

完整代码：

/*0ms,0KB*/

#include<cstdio>
#include<cstring>

char word[101];

int main()
{
	int n, len;
	scanf("%d", &n);
	getchar();///使用gets前一定要把'\n'读掉！
	while (n--)
	{
		memset(word, 0, sizeof(word));
		gets(word);
		len = strlen(word);
		if (len > 10)
			printf("%c%d%c\n", word[0], len - 2, word[len - 1]);
		else
			printf("%s\n", word);
	}
	return 0;
}