1031. Hello World for U (20)

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#C++ #PAT #Hello World for U

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本博客介绍了一种将任意长度不少于5个字符的字符串排列成U形的方法，包括输入输出规范、实现逻辑及代码示例。

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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 <iostream>
#include <cstdio>

using namespace std;

int main(){
	string s;
	cin >> s;

	int n = (int)s.size() + 2;
	int n1, n2, n3;
	int r = n % 3;

	if(r == 0){
		n1 = n2 = n3 = n / 3;
	}else{
		n2 = (n - r) / 3 + r;
		n1 = n3 = (n - r) / 3;
	}

	int i = 0, j = (int)s.size() - 1;
	for(; i < n1-1; ++i, --j){
		printf("%c", s[i]);
		for(int k = 0; k < n2-2; ++k)
			printf(" ");
		printf("%c\n", s[j]);
	}

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

	return 0;
}