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
—————————————————————————
- 由题意得,两侧边要尽可能大但是要小于底边,然后或几张图找规律可以得出。
- 可以直接打印图形也可以赋一个二维数组,这里是赋一个二维数组,先全部初始化为空格,然后再赋值侧边底边,最后打印。
#include<iostream>
#include<string>
using namespace std;
int main()
{
char ans[50][50]; string str;
getline(cin, str);
int n = str.length();
int n1 = (n + 2) / 3, n3 = n1, n2 = n + 2 - n1 - n3;
for (int i = 0; i < n1; i++)
{
for (int j = 0; j < n2; j++)
{
ans[i][j] = ' ';
}
}
int pos = 0;
for (int i = 0; i < n1; i++)
{
ans[i][0] = str[pos++];
}
for (int j = 1; j < n2 - 1; j++)
{
ans[n1 - 1][j] = str[pos++];
}
for (int i = n3 - 1; i >= 0; i--)
{
ans[i][n2-1] = str[pos++];
}
for (int i = 0; i < n1; i++)
{
for (int j = 0; j < n2; j++)
{
cout << ans[i][j];
}
cout << endl;
}
}