CF 71A [字符串统计]

比较水的一道题，不过WA了一次，没看清题意；

题意：一个单词，如果它的长度小于等于10，则输出这个单词，否则留头和尾然后中间输出省略字符的个数；

简单的方法就是求出长度len，然后把字符数组的第一个和最后一个输出一下就好了；中间加上len-2的个数就是答案；

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

4wordlocalizationinternationalizationpneumonoultramicroscopicsilicovolcanoconiosis

output

wordl10ni18np43s

#include <iostream>#include <string.h>#include <algorithm>#include <cstdio>using namespace std ;char a[200];int main(){int n ;cin>>n;while(n--){memset(a,0,sizeof(a));cin>>a;int len = strlen(a);if(len<=10){cout<<a<<endl;}else {cout<<a[0]<<len-2<<a[len-1]<<endl;}}return 0 ;}