Hello world for you

lowor
a m
c o
. c
jobdu.
参考答案：
/*先输入a[],
1.分配n1,n2,n3,n1=n3,n3=N+2-2*n1,且n2>=n1;
2.设置打印函数，打印n2-2个空格
3.打印a[0],n2-2个空格,a[j-1];
打印a[1],n2-2个空格,a[j-2];
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.
打印a[2],n2-2个空格,a[j-3];
...........................
打印a[n1-2],n2-2个空格,a[j-n1+1];
打印a[n1-1],a[n1],a[n1+1].......a[j-n1];
*/
#include<stdio.h>
}
for(i=n1-1;i<j-n1+1;i++)
printf("%c",a[i]);
printf("\n");
}
return 0;
} Biblioteka 浙大计算机复试编程题 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:
#include<string.h>
void show(int n)
{
int i;
for(i=0;i<n;i++)
printf(" ");
}
int main()
{
char a[80];
int n1, n2, i, j;
while(scanf("%s",a)!=EOF)
n2=j+2-2*n1;
for(i=0;i<n1-1;i++)
{
printf("%c",a[i]);
show(n2-2);
printf("%c\n",a[j-1-i]);
{
j = strlen(a);
for(n1=1,n2=j+2-2*n1;n2>=n1&&n2>=3;)
{
n1++;
n2=j+2-2*n1;
}
n1--;
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.
输出：
For each test case, print the input string in the shape of U as specified in the description.
样例输入：
helloworld!

样例输出：
h !
e d
l l