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
1
2
3
4

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!
1

Sample Output:

h   !
e   d
l   l
lowor
1
2
3
4

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
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;
}

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