题目链接：Problem - 1473B - Codeforces

解题思路：

先算出两个字符串的最小公倍数，再分别除以字符串长度，构造出两个新的字符串，就是最小公倍数除以老字符串长度个老字符串相加，打个比方，a字符串长度为4，b字符串长度为6，最小公倍数就是12，那么新字符串a就是12 / 4个老的字符串a相加，b同理，如果两个新的字符串相等，九叔出一个新的字符串，否则输出-1。

下面是c++代码：

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
    string boring(string, int);
    int lcm(int, int);
    int t, n;
    string a, b;
    cin >> t;
    while (t != 0) {
        cin >> a >> b;
        int alength = a.length(), blength = b.length();
        int g =  lcm(alength, blength);
        if (boring(a, g / alength) == boring(b,  g / blength)) {
            cout << boring(a, g / alength) << endl;
        }
        else {
            cout << -1 << endl;
        }
        t--;
    }
    return 0;
}
string boring(string s, int k) {
    string str = "";
    while (k != 0) {
        str += s;
        k--;
    }
    return str;
}
int lcm(int x, int y)
{
    if (x > y) {
        x = x ^ y;
        y = x ^ y;
        x = x ^ y;
    }
    int num = x, count = 1;
    while (1) {
        num = x * count;
        if (num % x == 0 && num % y == 0) {
            return num;
        }
        count++;
    }
}