1031. Hello World for U (20)

最新推荐文章于 2020-04-09 13:41:38 发布

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本文介绍了一种将任意长度（大于等于5）的字符串按U形打印的方法，确保字符顺序不变，同时使得U形尽可能接近正方形。文章提供了一个具体的实现示例，包括输入输出规范及样例。

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

一开始看题目中上下空这么多还以为要有个空格呢，，其实并不用

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

void print(int n1, int n2, int n3, string word)
{
    int counter = 0;
    char graph[n1][n2];
    for(int i=0; i<n1; i++)
    {
        for(int j=0; j<n2; j++)
            graph[i][j] = ' ';
    }
    for(int i=0; i<n1; i++)
        graph[i][0] = word[counter++];
    for(int i=1; i<n2-1; i++)
        graph[n1-1][i] = word[counter++];
    for(int i=n1-1; i>=0; i--)
        graph[i][n2-1] = word[counter++];
    for(int i=0; i<n1; i++)
    {
        for(int j=0; j<n2; j++)
            cout << graph[i][j];
        cout << endl;
    }
}
int main()
{
    int n1, n2, n3;
    string word;
    cin >> word;
    for(n2=3; n2<word.size(); n2++)
    {
        for(n1=0; n1<n2+1; n1++)
        {
            if(n1*2+n2-2 == word.size())
                break;
        }
        if(n1 != n2+1)
            break;
    }
    n1 = n3 = (word.size()+2-n2)/2;
    print(n1, n2, n3, word);
    return 0;
}