1031. Hello World for U (20)

题目如下：

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.

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

题目的关键在于对n₁ = n₃ = max { k| k <= n₂ for all 3 <= n₂ <= N } with n₁ + n₂ + n₃ - 2 = N的把握，题目之前一句提到要让U的形状越方越好，通过这个算式我们知道n2的范围是3到N，n1=n3，我们知道，n2越大，n1和n3救会越小，要让n1和n3最大，必须找到最小的满足条件的n2，这里的一个隐含条件是N + 2 - n2为偶数，也就是n1=n3所带来的条件，因此我们让n2从3到N变化，如果得到的N+2-n2是偶数，则再判断是否它的一半（k）满足小于等于n2，如果满足则已经找到合适的k，否则继续寻找，具体实现如下：

    for(n2 = 3; n2 <= N; ++n2){
        int temp = N + 2 -n2;
        if(temp%2 != 0)
           continue;
        n1 = temp/2;
        if(n1 <= n2)
           break;
    }

在找到了k值之后，n1、n2、n3就确定了，下面只需要把输入的字符依次压入n1、n2、n3容器，然后按照题目的规则输出即可，这里使用的是vector<char>来容纳字符，具体代码如下：

#include <iostream>
#include <string.h>
#include <string>

using namespace std;

int main()
{
    string input;
    string str1 = "";
    string str2 = "";
    string str3 = "";
    int n1,n2,n3;
    int temp = 0;
    cin >> input;
    int N = input.length();
    for(n2 = 3; n2 <= N; ++n2)
    {
        int temp = N + 2 -n2;
        if(temp%2 != 0)
                continue;
        n1 = temp/2;
        if(n1 <= n2)
            break;
    }
    n3 = n1;
    int i = 0;
    int offset = 0;
    for(i = 0; i < n1 - 1; i++){
        str1.push_back(input[i + offset]);
    }
    offset += i;
    for(i = 0; i < n2; i++){
        str2.push_back(input[i + offset]);
    }
    offset += i;
    for(i = 0; i < n3 - 1; i++){
        str3.insert(str3.begin(),input[i + offset]);
    }

    for(int i = 0; i < n1 -1 ;i++){
        cout << str1[i];
        for(int j = 0; j < n2 -2; j++) cout << " ";
        cout << str3[i] << endl;
    }
    for(int i = 0; i < n2; i++){
        cout << str2[i];
    }
    cout << endl;

    return 0;
}