1031. Hello World for U (20)题解

最新推荐文章于 2021-12-06 16:10:36 发布

原创最新推荐文章于 2021-12-06 16:10:36 发布 · 527 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#pat

pat 专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一种将任意长度的字符串排列成U形的方法，并通过示例解释了如何确保U形尽可能接近正方形，同时提供了完整的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题意就是生成一个u，根据要求n₁ = n₃ = max { k| k <= n₂ for all 3 <= n₂ <= N } with n₁ + n₂ + n₃ - 2 = N.可以看出，它要求这个u尽量是均匀的。于是我们知道 n2和n3还有n1平均分的时候是最均匀的，但是还是有不能平均分的情况（只会多出1个字符），因为n1 n3是对称的，所以需要只能给n2增加一个长度

代码：

#include <stdio.h>
#include <string.h>
using namespace std;
int a[100][100];
int main()
{
   int n;
   char s[100];
   gets(s);
   n=strlen(s);
   int n2=(n-2)/3;//获取n2,这里的n2不包括左右两端点

   int n1=(n-n2)/2;
    if(n1+n1+n2!=n)n2++;//如果没办法平均分 则让n2的长度增加1个
   int k=0;

//第一列
   for(int i=0;i<n1;i++)
   {
       a[i][0]=s[k];
       k++;
   }

//最下面一行
   for(int i=1;i<=n2;i++)
   {
       a[n1-1][i]=s[k];
       k++;
   }

//最右边一列
   for(int i=n1-1;i>=0;i--)
   {
       a[i][n2+1]=s[k];
       k++;
   }

//输出
   for(int i=0;i<n1;i++)
   {
       for(int j=0;j<n2+2;j++)
       {
           if(a[i][j]==0)printf(" ");
           else printf("%c",a[i][j]);
       }
       if(i!=n1-1)
       printf("\n");
   }

}