思路分析：
由题意得n1,n2,n3分别为U左侧、底部和右侧字母数，满足
n1+n2+n3=N（字符串长度）条件1
3=
#include"stdafx.h"
#include<string.h>
int main()
{int n1=0, n2, n;char str[85];scanf("%s", str);n = strlen(str);n2 = n;while (n2 >= 3 && n2 <= n && n1<=n2) {n1++;n2 = n + 2 - 2 * n1;}n1--;n2 = n + 2 - 2 * n1;for (int i = 0; i < n1-1; i++) {printf("%c", str[i]);for (int j = 0; j < n2 - 2; j++) {printf(" ");}printf("%c\n", str[n -1- i]);}for (int i = n1 - 1; i < n2 + n1 - 1; i++) {printf("%c", str[i]);}return 0;
}题目要求：
1031 Hello World for U (20)（20 分）
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