PatA1031：Hello World for U-优快云博客

本文介绍了一种将任意长度字符串转换为U形排列的算法，适用于至少5个字符的字符串。该算法首先从左垂直线开始向下打印字符，然后沿底部线从左向右打印，最后沿右垂直线向上打印，确保U形尽可能接近正方形。

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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.

输入格式

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.

输出格式

For each test case, print the input string in the shape of U as specified in the description.

输入样式

helloworld!

输出样式

h   !
e   d
l   l
lowor

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <sstream>
#include <cmath>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <map>
using namespace std; 

int main(){ 

  char str[100];

  scanf("%s", &str);

  int N = strlen(str), i, j;

  int n1 = (N + 2) / 3, n3 = n1, n2 = N + 2 - n1 - n3;

  for(i = 0; i < n1 - 1; i ++){

	  printf("%c", str[i]);
	   
	  for(j = 0; j < n2 - 2; j++){
	      
	  	printf(" ");
	  
	 	printf("%c\n", str[N - i - 1]);
	
	   }
	   
  }
	    
  for(i = 0; i < n2; i++){
       
  	printf("%c", str[n1 + i - 1]);

  }

  return 0;

}