PAT甲級 A1031

題目情況

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

解題思路

在本題中，需要根據(jù)題目的需要分析題目并且得到每一列和每一行所需要輸出的字符的個數(shù)。在上述的要求中，要求k≤n?2并且n?1??+n?2??+n?3???2=N，我們聯(lián)立以上不等式和等式即可得到當前的K的限制：k≤(n+2)/3,按照此思路得到k即可進行輸入：

#include<iostream> #include<string> #include<vector> #include<cmath> #include<string> #include<algorithm> using namespace std; int main() {string a; cin >> a;int n = a.size();int k = (n + 2) / 3;int n2 = n + 2 - k - k;int out = n2 - 2;for (int i = 0; i < k; i++) {cout << a[i];if (i == k - 1) {for (int j = 1; j <= out; j++) {cout << a[i + j];}}else {for (int j = 0; j < out; j++) {cout << " ";}}cout << a[a.size() - 1 - i];cout << '\n';} }

總結(jié)

以上是生活随笔為你收集整理的PAT甲级 A1031的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。