PAT 甲 1031 U型 Hello World for U

最新推荐文章于 2021-08-18 19:29:53 发布

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#字符串 #c++ #算法

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本文介绍了一种将任意长度的字符串以U形排列的算法，通过确定垂直和底部行的字符数，使U形尽可能接近正方形。文章提供了C++实现代码，用于按特定规则输出字符串。

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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≤n 2 ≤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

题意:
给定一个长度为 N 的字符串，请你将它以 U 形输出。
例如，helloworld 可以输出为：

h  d
e  l
l  r
lowo

也就是说，必须按照原始顺序输出字符，左垂直线自上而下共有 n1 个字符，底部行从左到右共有 n2 个字符，右垂直线自下而上共有 n3 个字符。
另外，必须满足 n1=n3=max{k|k≤n2对于所有3≤n2≤N} 以及 n1+n2+n3−2=N。
对于 helloworld，n1=n2=n3=4。
思路:
找规律即可，左右两侧的数量n1 = n3 = (len + 2)/3 n2=len+2-n1-n3
中间空格数量：n2-2
最后一行定位字母位置：s[n1-i-1]
C++:

#include<bits/stdc++.h>
using namespace std;
int main(){
    string s;
    cin>>s;
    int len = s.size();
    for( int i = 0 ; i < (len+2)/3-1 ; i++){
        cout<<s[i];
        for( int j = 0 ; j<len+2-(len+2)/3-(len+2)/3 -2 ; j++) cout<<" ";
        cout<<s[len-i-1]<<endl;
    }
    for(int i = 0 ;i<len+2-(len+2)/3-(len+2)/3 ; i++){
        cout<<s[(len+2)/3+i-1];
    }
    return 0;
}