A1031 Hello World for U (20分)

1031 Hello World for U (20分)

大佬的地址：刷PTA可以学习的大佬的地址

https://blog.csdn.net/liuchuo 婼神的地址

https://www.mmuaa.com/post/category/pat 斐斐のBlog

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

生词

vertical 垂直的

left to right along the bottom 从 左到右到底部

题目大意

给你一个字符串，要求输出一个U字形 底部的长度要不低于(可以等于)边部

分析（我的脑瘫分析）

① 无脑写两个循环,求出最长的边

② 根据边算出底部,循环输出

代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main(void) {
    char Array[81];
    int n1=0, n2, N,i;
    scanf("%s", &Array);
    N = strlen(Array);
    for (i = 1; i < N; i++) {
        for (int j = 1; 2 * i + j <= 2 + N; j++) {
            if (j >= i) {
                if (i >n1) {
                    n1 = i;
                    n2 = j;
                }
            }
        }
    }
    for (i = 0; i < n1-1; i++) {
        printf("%c", Array[i]);
        for (int j = 0; j < N-2*n1; j++) {
            printf(" ");
        }
        printf("%c\n", Array[N-i-1]);
    }
    int T=N-i-1;
    for (; i <=T; i++) {
        printf("%c", Array[i]);
    }
    
}

下面是看了大佬的博客之后发现的

下面贴一个 斐斐のBlog(她博客就叫这个)---侵权删 的代码

大佬的简洁的代码

#include <iostream>
#include <string>
using namespace std;

int main(){
    string input;
    cin >> input;
    int len = input.length();
    int side = (len + 2) / 3;
    int buttom = len + 2 - (2 * side);
    for(int i = 0; i < side - 1; i++){
        cout << input[i];
        for(int j = 0; j < buttom - 2; j++) cout << ' ';
        cout << input[len - 1 - i];
        cout << endl;
    }
    for(int i = side - 1; i < side + buttom - 1; i++)
        cout << input[i];
    cout << endl;
    return 0;
}