2012浙大复试上机题Helloworld_已知程序段s=hello world a=len-优快云博客

本文链接：https://blog.youkuaiyun.com/suichen1/article/details/8693746

本教程展示了如何将任意长度大于等于5的字符串以U形布局打印，同时确保布局尽可能方正，实现从上到下垂直打印n1个字符，水平打印n2个字符，最后从下到上垂直打印n3个字符。

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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.

输入：

There are multiple test cases.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.

输出：

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

样例输入：

helloworld!
ac.jobdu.com

样例输出：

h   !
e   d
l   l
lowor
a    m
c    o
.    c
jobdu.#include <iostream>
#include <string>
using namespace std;

int main()
{
	string s;
	int len;
	int i, j;

	while(cin >> s)
	{
		len = s.length() + 2;

		if(len % 3 == 0)
		{
			for(i = 0; i < len/3; i++)
			{
				cout << s[i];
				if(i != len/3 - 1)
				{
					for(j = 0; j < len / 3 - 2; j++)
						cout << " ";
					cout << s[s.length()-1-i] << endl;
				}
			}

			for(i = len / 3; i < 2 * (len/3) - 1; i++)
				cout << s[i];
		}
		else if(len % 3 == 1)
		{
			for(i = 0; i < len/3; i++)
			{
				cout << s[i];
				if(i != len/3 - 1)
				{
					for(j = 0; j < len / 3 - 1; j++)
						cout << " ";
					cout << s[s.length()-1-i] << endl;
				}
			}

			for(i = len / 3; i < 2 * (len/3); i++)
				cout << s[i];
		}
		else
		{
			for(i = 0; i < len/3; i++)
			{
				cout << s[i];
				if(i != len/3 - 1)
				{
					for(j = 0; j < len / 3; j++)
						cout << " ";
					cout << s[s.length()-1-i] << endl;
				}
			}

			for(i = len / 3; i < 2 * (len/3) + 1; i++)
				cout << s[i];
		}
		cout << endl;
	}
	return 0;
}