1031. Hello World for U (20)

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本文介绍了一种将任意长度大于等于5的字符串以U形排列的算法实现，该算法需确保形成的U形状尽可能接近正方形，并按原字符顺序打印。通过计算行列数，实现了字符串的有效布局。

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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

这题要把长度加2在进行计算，因为根据题意左下角和右下角的数被用了两次

行数=len/3

列=行数+len%3

#include<iostream>  
#include<cstring>  
#include<cstdio>  
#include<queue>  
#include<stack>  
#include<algorithm>  
#include<vector> 
#include<set>
using namespace std;
int main(){
	char s[100];
	gets(s);
	char out[100][100];
	for(int i=0;i<100;i++) for(int j=0;j<100;j++) out[i][j]=' ';
	
	int len=strlen(s)+2;
	int n=len/3;
	int m=n+len%3;
	int cnt=0;
	for(int i=0;i<n;i++)
		out[i][0]=s[cnt++];
	for(int i=1;i<m;i++)
	    out[n-1][i]=s[cnt++];
	for(int i=n-2;i>=0;i--)
	    out[i][m-1]=s[cnt++];
	    
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++)
		cout<<out[i][j];
		cout<<endl;
	}
	return 0;
}