[PAT]1031. Hello World for U (20)（Java实现）

1031. Hello World for U (20)

时间限制

400 ms

内存限制

65536 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  de  ll  rlowo

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n₁characters, 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   dl   llowor

因为字符串长度不超过80，所以n1、n2、n3的长度可以用暴力解法求解

package go.jacob.day1128;import java.util.Scanner;/** *  * @author Administrator * 1031. Hello World for U (20) */public class Demo2 {public static void main(String[] args) {Scanner sc = new Scanner(System.in);String s = sc.next();int num = s.length();int n1, n3 = num;for (n1 = 2; n1 < num; n1++) {n3 = num + 2 - 2 * n1;if (n3 < n1) {n1--;n3 = num + 2 - 2 * n1;break;}}int head = 0, tail = num - 1;for (int i = 0; i < n1 - 1; i++) {for (int j = 0; j < n3; j++) {if (j == 0)System.out.print(s.charAt(head++));else if (j == n3 - 1)System.out.print(s.charAt(tail--));elseSystem.out.print(' ');}System.out.println();}for (int j = 0; j < n3; j++) {System.out.print(s.charAt(head++));}sc.close();}}