Hello World for U 比较简单的思路，只需要一个string类型就可以了。

我想的很简单，就是按照一步步输出就可以了。没有用到什么二维数组，只要要个string类型就可以。

题目描述

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.

输入描述:

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.

输出描述:

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

示例1

输入

复制

helloworld!
www.nowcoder.com

输出

复制

h   !
e   d
l   l
lowor

我的AC代码

#include <bits/stdc++.h>
using namespace std;     //“百万”开头

int main()
{
    string str;
    while(cin >> str)
    {
        int len = str.length();
        int n = 2;               //标记竖起来的字符个数，初始是首尾两个字符竖起来
        while(len-n >= (n/2+1))  //当下面的字符个数 >= 单边竖起来的字符个数
            n += 2;              //两边继续各竖起来一个
        n -= 2;                  //最后会多减一次，要减掉
        for(int i = 0; i < n/2; i++)
        {
            int j = len - i - 1;   //j标记从末尾开始的字符下标
            printf("%c", str[i]);
            for(int k = 0; k < len - n - 2; k++)   //空格个数为 最后一行的字符个数-2
                printf(" ");
            printf("%c\n",str[j]);
        }
        //输出最后一行
        for(int k = n/2; k < len - n/2; k++)
            printf("%c", str[k]);
        printf("\n");
    }
    return 0;
}