A1031Hello World for U

Given any string of N (≥) characters, you are asked to form the characters into the shape of U. For example, helloworld can be printed as:

h  d
e  l
l  r
lowo

 

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n​1​​characters, then left to right along the bottom line with n​2​​ characters, and finally bottom-up along the vertical line with n​3​​ characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n​1​​=n​3​​=max { k | k≤n​2​​ for all 3 } with n​1​​+n​2​​+n​3​​−2=N.

Input Specification:

Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

Output Specification:

For each test case, print the input string in the shape of U as specified in the description.

Sample Input:

helloworld!

 

Sample Output:

h   !
e   d
l   l
lowor

思路：

•要求n1=n3<=n2,又因为n1+n2+n3-2=N 其中是n1尽可能大，所以有三种情况

1. 如果n % 3 == 0，n正好被3整除，直接n1 == n2 == n3;

2. 如果n % 3 == 1，因为n2要⽐n1⼤，所以把多出来的那1个给n2

3. 如果n % 3 == 2, 就把多出来的那2个给n2

所以得到公式：n1 = n / 3，n2 = n / 3 + n % 3。

•又因为没有办法按行列顺序输出，所以把他们存储到二维字符数组中，一开始初始化字符数组为空格，然后按u型填充进去，最后打印二维数组。

以上参考柳神笔记和我自己的理解(*/ω＼*)

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
int main() {
    string str;
    cin >> str;
    int n = str.length() + 2;
    int n1 = n / 3, n2 = n / 3 + n % 3;
    int index = 0;
    char a[30][30];
    memset(a, ' ', sizeof(a));
    for (int i = 0; i < n1; i++)a[i][0] = str[index++];
    for (int i = 1; i < n2 - 1; i++)a[n1 - 1][i] = str[index++];
    for (int i = n1 - 1; i >= 0; i--)a[i][n2 - 1] = str[index++];
    for (int i = 0; i < n1; i++) {
        for (int j = 0; j < n2; j++) {
            cout << a[i][j];
        }
        cout << endl;
    }
    return 0;
}