Hello World for U

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.

题目本身不难，在做的过程中遇到了三个问题

1.由于第一次尝试用c++写，虽然跟c语言相差无几，但是有需要注意的细节。用到了string类，需要引入cstring包，但是VC引入string包才能编译通过，在OJ上只能是ctring才能编译通过

2.由于n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N }，也就是说n1,n2,n3三个数可以相等，所以在判断的时候要加n1<=n2

3.又由于n2>=3的，最开始做的时候没看到等号，直接让n2从4开始的，导致当n=5的时候出现了错误

修正三个错误后的代码如下：（其中n1,n2分别表示了题目中的n1,n3;x表示题目中的n2）

#include <iostream>
#include <cstring>
#inculde <cstdio>

using namespace std;
int main(){
    char arr[80];
    int x;
    int i,j;
    int n1,n2;
    while(scanf("%s",arr)!=EOF){
    int n = strlen(arr);
    for(x=3;x<n;x++){
        if((n+2-x)%2==0&&(n+2-x)/2<=x)
        break;               
    }
    n1=n2=(n+2-x)/2;
    for(i=0;i<n1-1;i++)
    {
        cout<<arr[i];
        for(j=0;j<x-2;j++)
            cout<<" ";
        cout<<arr[n-i-1]; 
        cout<<"
";    
    }
    for(i=0;i<x;i++)
    {
        cout<<arr[n1-1+i];
    }
    cout<<"
";    
    }
    return 0;
}