PAT甲级真题 1031 Hello World for U (20分)（数学方法计算出n1、n2）

最新推荐文章于 2021-08-13 20:15:05 发布

原创最新推荐文章于 2021-08-13 20:15:05 发布 · 303 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#字符串 #算法

PAT 专栏收录该内容

209 篇文章

订阅专栏

本文介绍了一种将任意长度的字符串转换成U形排列的方法，通过计算确定n1和n2的值，使得字符串在垂直和底部水平方向上均匀分布，同时保持尽可能接近正方形的形状。文章详细解释了算法思路，并提供了完整的C++代码实现。

题目

Given any string of N (≥) 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 n1characters, 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 } 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

思路

关键就是求出n1、n2。

n1=n3=max {k|k≤n2 for all 3}这个条件有点难懂，可以理解为n1=n3<=n2。

由题目有：

n1 <= n2 = N - 2 * n1 + 2

n1 <= (N + 2) / 3

所以可直接令int类型的n1 = (N + 2) / 3，再按要求打印字符即可：

前n1-1行，每行打印首尾对应的字符，中间包含n2-2个空格；最后一行，从第n1个字符开始，连续打印n2个字符。

代码

#include <iostream>
using namespace std;

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