九度OJ 题目1464：Hello World for U

一.题目描述：

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.

输入：

There are multiple test cases.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.

输出：

For each test case, print the input string in the shape of U as specified in the description.

样例输入：

helloworld!

ac.jobdu.com

样例输出：

h !

e d

l l

lowor

a m

c o

. c

jobdu.

二.题目分析

本题中算好n1,n2,n3的关系，n1=n3,n1+n2+n2-2=N,n2=(n+2)/3即可

三.代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 82
int main()
{
char str[MAX];
int n1,n2,len,i,j1,j2,k;

while(scanf("%s",str)!=EOF)
{
len=strlen(str);
n1=(len+2)/3;
n2=len+2-2*n1;
j1=0;
j2=len-1;

for(i=0;i<n1;i++)
{
if(i!=n1-1)
{
for(k=0;k<n2;k++)
{
if(k==0)
{
printf("%c",str[j1]);
j1++;
}

else if(k==n2-1)
{
printf("%c\n",str[j2]);
j2--;
}

else
printf(" ");
}
}
else
{
for(k=j1;k<=j2;k++)
printf("%c",str[k]);
printf("\n");
}
}
}

return 0;
}