# 题目

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.
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

1.n1 + n2为偶数
2.n1 <= n2

# 源代码

#include <iostream>
#include <string>

using namespace std;

int main(){
int n1, n2;
string s;
cin >> s;
int n = s.size();
// 先求出行数n1和列数n2
for (n2 = 3; n2 <= n; ++n2){
int tmp = n + 2 - n2; // tmp = n1 + n3 = 2 * n1;
if (tmp % 2 != 0){
continue;
}
n1 = tmp / 2;
if (n1 <= n2){
break;
}
}
int blank = n2 - 2;
int row = n1 - 1;
for (int i = 0; i < n1; ++i){ // 行数
if (i < row){ // 前三行
printf("%c", s[i]);
cout << string(blank, ' ');
printf("%c\n", s[n - 1 - i]);
}
else{
for (int j = 0; j < n2; ++j){ // 最后一行
printf("%c", s[i + j]);
}
}
}
return 0;
}