点击链接PAT甲级-AC全解汇总
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 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
题意：
输入一串字符串，把他按照U型输出，要求是竖着的高度不超过水平的宽度。
我的思路：
题目中的k是竖着的高度，最低下一行也算（所以k=4，n2=5），我这里方便计算，k是去掉最下面一行的高度，所以k=3，n2=5 。
经过数学计算，很容易得出对于长度N，K应该是(N-1)/3，剩下的给n2；
所以：
k = (N - 1) / 3;
n2 = N - 2 * k;
知道竖着有k个 横着有n2个，那就一行行输出好了
竖着的部分第i行：就是正数和倒数第i个字符；
横着的部分，就是正好从k开始，共有n2个字符；
这就是为什么我算 k 的时候不算最后一行的原因！
我的代码：
#include<bits/stdc++.h>
using namespace std;int main()
{
string str;cin>>str;int N=str.length();int k=(N-1)/3;int n2=N-2*k;string blank(n2-2,' ');for(int i=0;i<k;i++)cout<<str[i]<<blank<<str[N-i-1]<<endl;cout<<str.substr(k,n2)<<endl;return 0;
}