[C++知识库]1031 Hello World for u

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

编译软件：visual studio

编译语言：c语言

参考代码：

#include<cstdio>
#include<cstring>

int main()
{
?? ?char str[100], ans[40][40];
?? ?gets_s(str, 100);
?? ?int N = strlen(str);
?? ?int n1 = (N + 2) / 3, n3 = n1, n2 = N + 2 - n1 - n3;
?? ?for (int i = 1; i <= n1; i++)
?? ?{
?? ??? ?for (int j = 1; j <= n2; j++)
?? ??? ?{
?? ??? ??? ?ans[i][j] = ' ';
?? ??? ?}
?? ?}
?? ?int pos = 0;
?? ?for (int i = 1; i <= n1; i++)
?? ?{
?? ??? ?ans[i][1] = str[pos++];
?? ?}
?? ?for (int j = 2; j <= n2; j++)
?? ?{
?? ??? ?ans[n1][j] = str[pos++];
?? ?}
?? ?for (int i = n3 - 1; i >= 1; i--)
?? ?{
?? ??? ?ans[i][n2] = str[pos++];
?? ?}
?? ?for (int i = 1; i <= n1; i++)
?? ?{
?? ??? ?for (int j = 1; j <= n2; j++)
?? ??? ?{
?? ??? ??? ?printf("%c", ans[i][j]);
?? ??? ?}
?? ??? ?printf("\n");
?? ?}
?? ?return 0;
}

注：在pat中gets_s(str,100)始终编译不出来，不知道是为什么，希望有懂的大佬能解答一下，我自己的处理方法就是将gets_s(str,100)改成了scanf("%s",str)