题目描述如下：

1031 Hello World for U (20 point(s))

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 n​1​​ characters, then left to right along the bottom line with n​2​​ characters, and finally bottom-up along the vertical line with n​3​​ characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n​1​​=n​3​​=max { k | k≤n​2​​ for all 3≤n​2​​≤N } with n​1​​+n​2​​+n​3​​−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

遇到的问题以及解决思路：

 题目的意思，n1和n3是两边的竖列，包括了最底下的字符，所以题目里会出现2，但是根据题目意思变通理解一下就好，保证最低行n2大于等于3，然后左右两列进行均等分配，如果出现两列的数量要大于底行n2了，那么再各自减去一个就好。

注意点！！！

c11标准里不再有gets()，然后可以使用fgets，但是要注意换行符也会被读入。

#include <cstdio>
#include <cstring>
//使用strlen
int main()
{
    int lenth;
    char strs[100];
    //gets(strs);
    //不要再用gets
    fgets(strs, sizeof(strs), stdin);
    //fgets会把换行符也读进去
    //puts(strs);
    lenth = strlen(strs) - 1;
    int tempcol = (lenth - 3) / 2;
    int temprow = lenth - 2 * tempcol;
    while (tempcol >= temprow) {
        tempcol--;
        temprow += 2;
    }
    int n = 0;
    for (int i = 1; i <= tempcol; i++) {
        printf("%c", strs[n]);
        for (int a = 1; a <= temprow - 2; a++) {
            printf(" ");
        }
        printf("%c\n", strs[lenth - i]);
        n++;
    }
    for(int j = 1; j <= temprow; j++) {
        printf("%c", strs[n++]);
    }

    printf("\n");
    return 0;
}