PAT 1031 Hello World for U[一般]

1031 Hello World for U （20 分）

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 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≤n​2​​≤N } 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

题目大意：给出一个字符串，要求按U型输出n1,n3分别是左边和右边从第一个到最后一个的高度，n2是底层的长度，要求满足n1=n3(k|k<=n2)，也就是列的高度要小于等于底部的长度。

//猛一看这个题目还挺难的，我的AC：

#include <iostream>
#include <vector>
#include<cstdio>
using namespace std;
 
int main() {
    string s;
    cin>>s;
    int size=s.size();
    int n13,n2;
    n13=size/;
    n2=n13+size%;
    if(n13==n2){//如果两者相等的话，那么需要进行修改。
        n13-=;
        n2=size-n13*;
    }
 
    for(int i=;i<size;i++){
        if(i<n13){
            cout<<s[i];
            for(int j=;j<n2-;j++){
                cout<<' ';
            }
            cout<<s[size--i]<<'\n';
        }else{
            break;
        }
 
    }
    //打印n2层
    for(int i=n13;i<size-n13;i++){
        cout<<s[i];
    }
    return ;
}

１.第一次提交时没通过，因为是如果是n13和n2相等的话，中间的空格就没办法输出了，所以就又添加了一个if判断是否相等，如果相等的话，那么就进行修改，便通过了。