1031 Hello World for U

1031 Hello World for U （20 分）

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​1​​ characters, 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

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题型分类：快乐模拟

题目大意：将给定的字符串输出成“U”型，并且使这个“U”尽可能的方。

解题思路：先解出最大的n1，接下来按照题意输出。

---

#include <iostream>
#include <string>

using namespace std;

int main(int argc, char** argv) {
	string str;
	cin >> str;
	int len = str.length();
	int n1 = 0, n2 = len;
	while(n1 <= n2 && n2 >= 3){ //不满足条件时退出 
		n1++;
		n2 = len + 2 - 2 * n1;
	}
	n1--; //回退一步
	n2 = len + 2 - 2 * n1;
	for(int i = 0; i < n1 - 1; i++){ //按行输出，先输出n1 - 1行 
		printf("%c", str[i]);
		for(int j = 0; j < n2 - 2; j++){ //一行的长度为n2，那么空格的个数为n2-2 
			printf(" ");
		}
		printf("%c\n", str[len - 1 - i]); 
	}
	for(int i = 0; i < n2; i++){ //输出最后一行 
		printf("%c", str[n1 - 1 + i]);
	}
	return 0;
}