1031 Hello World for U （不等式计算）

1031 Hello World for U （20 分）

Given any string of NNN (≥5ge 5≥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 n1n_1n​1​​ characters, then left to right along the bottom line with n2n_2n​2​​ characters, and finally bottom-up along the vertical line with n3n_3n​3​​ characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n1=n3=maxn_1 = n_3 = maxn​1​​=n​3​​=max { kkk | k≤n2k le n_2k≤n​2​​ for all 3≤n2≤N3 le n_2 le N3≤n​2​​≤N } with n1+n2+n3−2=Nn_1 + n_2 + n_3 - 2 = Nn​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

思路：
　　题目要求输出的图形要尽可能的方，并且限制了n1和n3要小于n2,且有n1+n2+n3-2=N，所以可知n1=（N+2）/3，n2=N-n1*2,n3=n1

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<string>
using namespace std;

int main()
{
    string str;
    cin>>str;
    int len=str.size()+2;
    int n1=len/3-1;
    int n2=len-(n1+1)*2;
    for(int i=0;i<n1;i++)
    {
        cout<<str[i];
        for(int j=0;j<n2-2;j++)
            cout<<" ";
        cout<<str[str.size()-i-1]<<endl;
    }
    for(int i=n1;i<n1+n2;i++)
        cout<<str[i];
    return 0;
}