PAT甲级 1031 Hello World for U (20 分)

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本文探讨如何将给定的至少5个字符的字符串转换成U形布局，遵循特定的打印顺序和形状要求，同时尽可能使U形面积最大化。实现算法并提供代码示例，适用于编程爱好者和字符串处理实践者。

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 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   d
l   l
lowor

#include <iostream>


using namespace std;


int main() {
    string n;
    while (cin >> n) {
        int N = n.size() + 2;
        int n1 = N / 3;
        int n2 = N - n1 * 2;
        for (int i = 0; i < n1; ++i) {
            for (int j = 0; j < n2; ++j) {
                if (!j and i != n1 - 1)
                    cout << n[i];
                else if (j == n2 - 1 and i != n1 - 1)
                    cout << n[n.size() - 1 - i];
                else if (i != n1 - 1)
                    cout << " ";
                if (i == n1 - 1)
                    cout << n[i + j];
            }
            cout << endl;
        }
    }
    return 0;
}

第二次刷此题代码

#include <iostream>


using namespace std;

int main() {
    string a;
    cin >> a;
    int n1 = (a.size() + 2) / 3;
    int n2 = a.size() - 2 * n1 + 2;
    int i;
    for (i = 0; i < n1 - 1; ++i) {
        cout << a[i];
        for (int j = 0; j < n2 - 2; ++j) {
            cout << " ";
        }
        cout << a[a.size() - 1 - i] << endl;
    }
    for (int j = 0; j < n2; ++j) {
        cout << a[i + j];
    }

    return 0;
}