PAT甲级 1031. Hello World for U (20)

题目：

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 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   dl   llowor

思路：

这道题属于运筹学的内容，有段时间没接触了，所以这里用了略笨的方法。

1.先根据3 <= n₂ <= N 的条件，遍历n2；

2.再根据n₁ + n₂ + n₃ - 2 = N和n₁ = n₃ 找到符合条件的值；

3.最后再根据条件k <= n₂和要求max { k| k <= n₂ for all 3 <= n₂ <= N } 筛选得到相应的n1，n2和n3.

得到相应的值后，再进行输出。这里还要注意的是处最后一行外，每一行都需要根据n2的值在字符之间输出n2-2个空格。

代码：

#include<iostream>#include<string>using namespace std;int main(){string S;cin >> S;int i;int n1, n2, n3;int k,N;n1 = 0;N = S.size();for (n2 = 3; n2 <= N; ++n2){if (!((N + 2 - n2) % 2)) //判断是否有符合等式限制条件的存在{k = (N + 2 - n2) / 2;if (k <= n2 && k > n1)//判断当前k是否满足条件且为最新找到的最大值n1 = k;}}n3 = n1;n2 = N + 2 - n1 - n3;//outputint j;for (i = 0; i < n1-1; ++i){cout << S[i];for (j = 0; j < n2 - 2; ++j)cout << " ";cout << S[N - 1 - i] << endl;}for (i = 0; i < n2; ++i){cout << S[n1 -1+ i];}cout << endl;system("pause");return 0;}