A1031 Hello World for U[图形输出]

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 n2 characters, 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 <= n2 for 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

—————————————————————————

1. 由题意得，两侧边要尽可能大但是要小于底边，然后或几张图找规律可以得出。
2. 可以直接打印图形也可以赋一个二维数组，这里是赋一个二维数组，先全部初始化为空格，然后再赋值侧边底边，最后打印。

#include<iostream>
#include<string>
using namespace std;
int main()
{
	char ans[50][50]; string str;
	getline(cin, str);
	int n = str.length();
	int n1 = (n + 2) / 3, n3 = n1, n2 = n + 2 - n1 - n3;
	for (int i = 0; i < n1; i++)
	{
		for (int j = 0; j < n2; j++)
		{
			ans[i][j] = ' ';
		}
	}
	int pos = 0;
	for (int i = 0; i < n1; i++)
	{
		ans[i][0] = str[pos++];
	}
	for (int j = 1; j < n2 - 1; j++)
	{
		ans[n1 - 1][j] = str[pos++];
	}
	for (int i = n3 - 1; i >= 0; i--)
	{
		ans[i][n2-1] = str[pos++];
	}
	for (int i = 0; i < n1; i++)
	{
		for (int j = 0; j < n2; j++)
		{
			cout << ans[i][j];
		}
		cout << endl;
	}
}