A14000. String LCM
编程题
普及/提高-
知识点
题目描述
Let's define a multiplication operation between a string $a$ and a positive integer $x$ : $a \cdot x$ is the string that is a result of writing $x$ copies of $a$ one after another. For example, "abc" $\cdot~2~=$ "abcabc", "a" $\cdot~5~=$ "aaaaa".
A string $a$ is divisible by another string $b$ if there exists an integer $x$ such that $b \cdot x = a$ . For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings $s$ and $t$ (defined as $LCM(s, t)$ ) is the shortest non-empty string that is divisible by both $s$ and $t$ .
You are given two strings $s$ and $t$ . Find $LCM(s, t)$ or report that it does not exist. It can be shown that if $LCM(s, t)$ exists, it is unique.
A string $a$ is divisible by another string $b$ if there exists an integer $x$ such that $b \cdot x = a$ . For example, "abababab" is divisible by "ab", but is not divisible by "ababab" or "aa".
LCM of two strings $s$ and $t$ (defined as $LCM(s, t)$ ) is the shortest non-empty string that is divisible by both $s$ and $t$ .
You are given two strings $s$ and $t$ . Find $LCM(s, t)$ or report that it does not exist. It can be shown that if $LCM(s, t)$ exists, it is unique.
输入格式
The first line contains one integer $q$ ( $1 \le q \le 2000$ ) — the number of test cases.
Each test case consists of two lines, containing strings $s$ and $t$ ( $1 \le |s|, |t| \le 20$ ). Each character in each of these strings is either 'a' or 'b'.
Each test case consists of two lines, containing strings $s$ and $t$ ( $1 \le |s|, |t| \le 20$ ). Each character in each of these strings is either 'a' or 'b'.
输出格式
For each test case, print $LCM(s, t)$ if it exists; otherwise, print -1. It can be shown that if $LCM(s, t)$ exists, it is unique.
输入输出样例
输入 #1
3 baba ba aa aaa aba ab
输出 #1
baba aaaaaa -1
说明/提示
In the first test case, "baba" = "baba" $\cdot~1~=$ "ba" $\cdot~2$ .
In the second test case, "aaaaaa" = "aa" $\cdot~3~=$ "aaa" $\cdot~2$ .
In the second test case, "aaaaaa" = "aa" $\cdot~3~=$ "aaa" $\cdot~2$ .