PAT 甲级 1031 Hello World for U (20 分)(一开始没看懂题意)

1031 Hello World for U (20 分)

 

Given any string of N (≥) 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 n​1​​ characters, then left to right along the bottom line with n​2​​ characters, and finally bottom-up along the vertical line with n​3​​ characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n​1​​=n​3​​=max { k | k≤n​2​​ for all 3 } with n​1​​+n​2​​+n​3​​−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：

一开始没看懂题意，看懂以后一遍过。

尽可能围成正方形。在满足n1=n3,且n1<=n2的条件下，n1尽可能地大。

底部字符数量 n2 最少为3。n1 = n3 要尽量等于 n2 。也就是说，n1 = n3 = （n+2） / 3；因为 '/ 3' 会截掉小数部分，所以 n1 和 n3 总是小于或者等于 n2 。

AC代码：

#include<bits/stdc++.h>
using namespace std;
char a[];
int main(){
    cin>>a;
    int l=strlen(a);
    int n1=(l+)/;
    int n2=l-n1*;
    //cout<<n1<<" "<<n2<<endl;
    for(int i=;i<n1-;i++){
        cout<<a[i];
        for(int j=;j<=n2;j++){
            cout<<" ";
        }
        cout<<a[l--i]<<endl;
    }
    for(int i=n1-;i<=n1+n2;i++){
        cout<<a[i];
    }
    return ;
}