pat 1031

1031. Hello World for U (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

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

char str[90];

void outPut(int n1, int n2, int n3)
{
	int i, j;
	int len = strlen(str) - 1;

	for(i = 0; i < n1 - 1; i++)
	{
		printf("%c", str[i]);

		for(j = 0; j < n2 - 2; j++)
			printf(" ");

		printf("%c\n", str[len--]);
	}

	for(; i < n1 + n2 - 1; i++)
		printf("%c", str[i]);

	printf("\n");
}

int main()
{
	int i, j;

	while(gets(str))
	{
		int n1, n2, n3;

		int len = strlen(str);
		len += 2;

		if(len % 3 == 0)
			n1 = n2 = n3 = len / 3;
		else if(len % 3 == 1)
		{
			n1 = n3 = (len-1) / 3;
			n2 = n1 + 1;
		}
		else if(len % 3 == 2)
		{
			n1 = n3 = (len - 2) / 3;
			n2 = n1 + 2;
		}

		outPut(n1, n2, n3);
	}
	return 0;
}