Zhejiang university----Hello World for U

//Zhejiang university----Hello World for U
/*
时间限制 1s 内存限制 128M

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.
*/
#include <iostream>
#include <string>
using namespace std;
void print(int n1,string str,int arrayLength,int n2){
    //打印结果
    int i,j;
    for(i=0;i<n1;i++){
        if(i==(n1-1)){
            cout<<str.at(i);
            for(j=i+1;j<arrayLength;j++){
                cout<<str.at(j);
            }
        }else{
            cout<<str.at(i);
            for(j=0;j<n2-2;j++){
                cout<< " ";
            }
            cout<<str.at(arrayLength - 1) <<endl;
            arrayLength--;
        }
    }
}
void getResult(int arrayLength,int &n1,int &n2){
    //进行求解
    if(arrayLength >= 5 && arrayLength < 80){
        //获取n2的最小结果
        int minResult = (arrayLength + 2)/3;
        if((arrayLength + 2)%3 == 0){
            n1 = n2 = minResult;
        }else{
            for(n2 = minResult;n2<arrayLength;n2++){
                for(n1=n2-1;n1>0;n1--){
                    if((2*n1+n2-2) == arrayLength){
                        //打印结果
                        return;
                    }
                }
            }
        }
    }
}
int main(){
    //用于记录求解结果
    int n1 = 0,n2 = 0;
    string str;
    //最小结果统计
    //数组长度
    int arrayLength;
    while(cin>>str){
        //cout<<str.length();
        //获得数组的长度
        arrayLength = str.length();
        //进行求解
        getResult(arrayLength,n1,n2);
        print(n1,str,arrayLength,n2);
        cout<<endl;
    }
    return 0;
}