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 lowoThat 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
题意就是生成一个u,根据要求n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.可以看出,它要求这个u尽量是均匀的。于是我们知道 n2和n3还有n1平均分的时候是最均匀的,但是还是有不能平均分的情况(只会多出1个字符),因为n1 n3是对称的,所以需要只能给n2增加一个长度
代码:
#include <stdio.h>
#include <string.h>
using namespace std;
int a[100][100];
int main()
{
int n;
char s[100];
gets(s);
n=strlen(s);
int n2=(n-2)/3;//获取n2,这里的n2不包括左右两端点
int n1=(n-n2)/2;
if(n1+n1+n2!=n)n2++;//如果没办法平均分 则让n2的长度增加1个
int k=0;
//第一列
for(int i=0;i<n1;i++)
{
a[i][0]=s[k];
k++;
}
//最下面一行
for(int i=1;i<=n2;i++)
{
a[n1-1][i]=s[k];
k++;
}
//最右边一列
for(int i=n1-1;i>=0;i--)
{
a[i][n2+1]=s[k];
k++;
}
//输出
for(int i=0;i<n1;i++)
{
for(int j=0;j<n2+2;j++)
{
if(a[i][j]==0)printf(" ");
else printf("%c",a[i][j]);
}
if(i!=n1-1)
printf("\n");
}
}