CodeForces - 1473B

CodeForces - 1473B

题目

Let’s define a multiplication operation between a string a and a positive integer x: a⋅x is the string that is a result of writing x copies of a one after another. For example, “abc” ⋅ 2 = “abcabc”, “a” ⋅ 5 = “aaaaa”.

A string a is divisible by another string b if there exists an integer x such that b⋅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.

Input
The first line contains one integer q (1≤q≤2000) — the number of test cases.

Each test case consists of two lines, containing strings s and t (1≤|s|,|t|≤20). Each character in each of these strings is either ‘a’ or ‘b’.

Output
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.
Example
inputCopy
3
baba
ba
aa
aaa
aba
ab
outputCopy
baba
aaaaaa
-1
Note
In the first test case, “baba” = “baba” ⋅ 1 = “ba” ⋅ 2.

In the second test case, “aaaaaa” = “aa” ⋅ 3 = “aaa” ⋅ 2.

题意：给定两个只含“a","b"字母的字符串，找到两个字符串共同的最小公倍序列。

思路：事实上，当有一组字符串长度为奇数时（除了1），无论如何都找不到LCM(a,b)。如果细分的话情况很多，比较复杂，观察到题目所给字符串长度<=20。这样的话其实不必啰嗦，直接对两个字符串暴力填补至齐长再对比是否相同即可，也不必担忧填补后超过20，因为那样必定有一个字符串长度为奇数，是不合法的。

code

#include <bits/stdc++.h>
using namespace std;
#define inf 1e5+10
#define ll long long
const int gm=10010;
string s1,s2;
ll q;
int main()
{
    cin>>q;
    while (q--)
    {
        cin>>s1>>s2;
        string temp1,temp2;
        temp1=s1;//存下s1,s2初始值，为了之后填补
        temp2=s2;
        while (s1.size()!=s2.size())
        {
            if (s1.size()<s2.size())
                s1+=temp1;//谁短补谁，最后总会相同
            else if (s1.size()>s2.size())
                s2+=temp2;
        }
        if (s1==s2)
            cout<<s1<<endl;
        else
        {
            cout<<-1<<endl;
        }
    }
    system("pause");
    return 0;
}

当然如果字符串长度增大，这个暴力算法就不适用了。