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：

**helloworld!

www.nowcoder.com**

> 输出：

h !

e d

l l

lowor

w m

w o

. .

n r

owcode

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    char a[80];
    while(scanf("%s", &a) != EOF){ //当不知道要输入多少次时用while，当知道循环次数时最好用for循环
        char marx[80][80];
        for(int i = 0; i < 80; i++){    //初始化数组
            for(int j = 0; j < 80; j++){
                marx[i][j] = ' ';
            }
        }
        
        int n = 0, k = 0 ,n1, n2; //此时的n1、n2代表数组中的下标，并不是长度，因为数组的下标从零开始
        while(a[n] != '\0') {  //计算出字符串的长度
            n++;
        }
        
        for(int i = 0; i < (n+2)/3; i++){ //给U型左边进行赋值
            marx[i][0] = a[k++];
            n1 = i;
        }
        
        for(int j = 1; j < (n+2)/3 + (n+2)%3 ; j++){ //给U型底进行赋值
            marx[n1][j] = a[k++];
            n2 = j;
        }
        
        for(int t = n1 - 1; t >= 0; t--){ //给U型右边进行赋值
            marx[t][n2] = a[k++];
        }
        
        for(int i = 0; i <= n1; i++){  //依次输出U型
            for(int j = 0; j <= n2; j++){
                printf("%c", marx[i][j]);
            }
            cout << endl;
        }
    }
    return 0;
}