技術標籤：演算法筆記

Hello World for U

題目描述
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:

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.

輸入
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.

輸出
For each test case, print the input string in the shape of U as specified in the description.

樣例輸入 Copy
helloworld!
樣例輸出 Copy

思路有了，看看具體的要求。字串的長度是N，n1，n3代表兩邊每列字元的數目。n2代表最後一行的字元數。題目中給了一個算式：

n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.

仔細研究這個算式，這裡的k是不大於n2的，也就是說n1和n3是不大於n2且滿足n1+n2+n3=N+2的最大值。那麼自然有n1=n3=(N+2)/3，n2=N+2-(n1+n3)。也就是說設side為兩邊的字元數（包括最後一行的兩端），則side=n1=n3=(N+2)/3。設mid為最後一行除去兩端的兩個字元後剩下的字元數，mid=N-side*2（總長度減去兩邊的字元數）。同時mid也是我們輸出除最後一行外前面所有行需要空出的空格數。

最後如何在第一行輸出第一個字元和最後一個字元呢？那自然是str[0]和str[len-1-i]（len為字串的長度，也就是N）。

於是問題完美解決，步驟如下：

1）計算字串長度len；

2）計算兩邊的字元數side=(len+2)/3；

3）計算最後一行中間的字元數（前面每行中間的空格數）；

4）輸出每行相應的字元。

程式碼

#include<cstdio>
#include<cstring>
const int maxn = 100;
int main(){
	char str[maxn];
	int len, side, mid;
	gets(str);
	len = strlen(str);
	side = (len + 2) / 3;
	mid = len - side * 2;
	int i = 0;
	for(int j = 1; j < side; j++){
		int t = mid;
		printf("%c", str[i]);
		while(t--){
			printf(" ");
		}
		printf("%c\n", str[len-1-i]);
		i++;
	}
	for(int k = i; k <= len-1-i; k++){ //最後一行 
		printf("%c", str[k]);
	}
	return 0;
}