1031. Hello World for U (20)

本文介绍了一种将任意长度大于等于5的字符串排列成U形的方法,并提供了一个具体的编程实现案例。该算法要求U形尽可能接近正方形,即左右两侧的字符数相同且等于底部边长的最大可能值。

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  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 n 1characters, then left to right along the bottom line with n 2 characters, and finally bottom-up along the vertical line with n 3 characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n 1 = n 3 = max { k| k <= n 2 for all 3 <= n 2 <= N } with n 1 + n 2 + n 3 - 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
 


   
  1. #pragma warning(disable:4996)
  2. #include <stdio.h>
  3. #include <string>
  4. #include <iostream>
  5. using namespace std;
  6. int main(void) {
  7. int n1, n2, n3;
  8. string s;
  9. cin >> s;
  10. n1 = n3 = (s.length() - 1) / 3;
  11. n1++; n3++;
  12. n2 = s.length() + 2 - 2 * n1;
  13. for (int i = 0; i < n1 - 1; i++) {
  14. cout << s[i];
  15. for (int i = 0; i < n2 - 2; i++)
  16. cout << ' ';
  17. cout << s[s.length() - i - 1] << endl;
  18. }
  19. for (int i = n1 - 1; i < n1 + n2-1; i++) {
  20. cout << s[i];
  21. }
  22. return 0;
  23. }





转载于:https://www.cnblogs.com/zzandliz/p/5023115.html

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