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题意：

Tom loves vowels, and he likes long words with many vowels. His favorite words are vowelly words. We say a word of length kk is vowelly if there are positive integers nn and mm such that n⋅m=kn⋅m=k and when the word is written by using nn rows and mm columns (the first row is filled first, then the second and so on, with each row filled from left to right), every vowel of the English alphabet appears at least once in every row and every column.

You are given an integer kk and you must either print a vowelly word of length kk or print −1−1 if no such word exists.

In this problem the vowels of the English alphabet are 'a', 'e', 'i', 'o' ,'u'.

 思路：

先求出k的所有因子，如果能找到两个大于等于5的因子，就挨个往矩阵里填因子，

否则就-1.

代码:

#include 
      
       using namespace std;typedef long long LL;char s[5] = {'a', 'e', 'i', 'o', 'u'};int main(){    int n;    cin >> n;    set
       
         yinzi;    for (int i = 1;i*i <= n;i++)        if (n%i == 0)            yinzi.insert(i), yinzi.insert(n/i);    for (auto it = yinzi.begin();it != yinzi.end();++it)    {        if ((*it) >= 5 && n/(*it) >= 5)        {            for (int i = 0;i < (*it);i++)            {                int pos = i;                for (int j = 1;j <= n/(*it);j++)                    cout << s[(pos++)%5];            }            return 0;        }    }    cout << -1 << endl;    return 0;}