PAT 1031 Hello World for U [图形打印]

最新推荐文章于 2020-06-21 11:04:00 发布

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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。算法目标是使U形尽可能接近正方形，通过合理分配左右垂直线和底部水平线的字符数实现。输入为一个不包含空格的字符串，输出为按U形排列的同一字符串。

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

枚举代码：

#include<stdio.h>
#include<string.h>

int main()
{
	char s[87];
	scanf("%s",&s);
	int len = strlen(s);
	int n1,n2,i,j;
	//因n1需要尽可能大，所以从小到大枚举,也可以从大到小加一个break
	for(i=0;i<=len;i++)
		if(i<=len+2-2*i) n1 = i; //满足n1<=n2 
	n2 = len+2-2*n1;
	//printf("%d %d\n",left,mid);
	int count = 0;
	for(i=0;i<len;i++)
	{
		printf("%c",s[i]);
		for(j=0;j<n2-2;j++)
		    printf(" ");
        printf("%c\n",s[len-i-1]);
        count++;
        if(count==n1-1) break;
	}
	for(i=n1-1;i<n1-1+n2;i++)
	    printf("%c",s[i]);
}

列方程得公式代码：

n1+n2+n3-2 = N ; n1<=n2 ; n1=n3 →n1<=(N+2)/3

#include<cstdio>
#include<cstring>

int main()
{
	int len,i,j,n1,n2,cnt = 0;
	char s[100];
	scanf("%s",s);
	len = strlen(s);
	n1 = (len+2)/3;
	n2 = len-2*n1+2;
	for(i=0;i<n1;i++)
	{
		if(i!=n1-1)
		{
			printf("%c",s[cnt++]);
			for(j=0;j<n2-2;j++) printf(" ");
			printf("%c",s[len-cnt]);
			printf("\n");
		}	
		if(i==n1-1)
		{
			for(j=0;j<n2;j++)
				printf("%c",s[cnt++]);
		}
	}
	return 0;
}