PAT 1031. Hello World for U (20)

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本文介绍了一种将任意长度大于等于5的字符串以尽可能接近正方形的U形格式输出的算法。通过确定合适的行数与列数比例，使得输出既符合原始字符顺序又保持U形的美观性。

Given any string of N (>=5) 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 n₁ characters, then left to right along the bottom line with n₂characters, and finally bottom-up along the vertical line with n₃ characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n₁ = n₃ = max { k| k <= n₂for all 3 <= n₂ <= N } with n₁ + n₂ + n₃ - 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
/*
 * 让输出一个字符串的U型输出。 保证U型字符串的行数是比列数小的最大值（枚举可得该值）。然后输出即可
*/
#include "iostream"
#include "string"
using namespace std;
int main() {
	string s;
	int col, row=0;
	cin >> s;
	int i, j;
	for(i = s.length();i>=3;i--)
		for (j = i;j >= 0; j--) {
			if (j > row && i+(j-1)*2==s.length()) {
				row = j;
			}
		}
	col = s.length() - (row-1) * 2;
	int k = 0;
	for (int i = 0; i < row; i++) {
		if (i != row - 1) {
			for (int j = 0; j < col; j++) {
				if (j == 0) {
					cout << s[k];
				}
				else if (j == col - 1)
					cout << s[s.length() - k - 1];
				else
					cout << " ";
			}
			k++;
		}
		else {
			for (int j = 0; j < col; j++)
				cout << s[k++];
		}
		cout << endl;
	}
	return 0;
}