版权声明:本文为博主原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明。
本文链接:https://blog.csdn.net/weixin_42449444/article/details/89435176
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.
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.
For each test case, print the input string in the shape of U as specified in the description.
helloworld!
h !
e d
l l
lowor
膜柳神~ 真的大佬(博文链接:https://www.liuchuo.net/archives/2053),她的思路是:假设n = 字符串长度 + 2,因为2 * n1 + n2 = n,且要保证n2 >= n1, n1尽可能地大,分类讨论:① 如果n % 3 == 0,n正好被3整除,直接n1 == n2 == n3; ② 如果n % 3 == 1,因为n2要比n1大,所以把多出来的那1个给n2;③ 如果n % 3 == 2, 就把多出来的那2个给n2;所以得到公式:n1 = n / 3,n2 = n / 3 + n % 3。初始化char型数组为空格,然后将字符串按照u型填充进去,最后输出这个数组u即可。
#include <bits/stdc++.h>
using namespace std;
int main()
{
string str;
getline(cin,str);
int n = str.length() + 2;
int n1 = n/3; //n1是左右两条竖线从上到下的字符个数
int n2 = n1 + n%3; //n2是底部横线从左到右的字符个数
char arr[30][30];
memset(arr,' ',sizeof(arr));
int cnt = 0; //用来记录当前字符的下标
for(int i = 0; i < n1; i++) //先排列左边那条竖线的n1个字符 hell
{
arr[i][0] = str[cnt++];
}
for(int i = 1; i < n2-1; i++) //再排列底部那条横线(除去最左最右字符外)的那n2-2个字符 owo
{
arr[n1-1][i] = str[cnt++];
}
for(int i = n1-1; i >= 0; i--) //最后排列右边那条竖线的n1个字符 rld!
{
arr[i][n2-1] = str[cnt++];
}
//打印U型图案
for(int i = 0; i < n1; i++)
{
for (int j = 0; j < n2; j++)
{
cout << arr[i][j];
}
cout << endl;
}
return 0;
}