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
这题实际上不难，就是刚开始要求没看明白，首先要尽量使宽与高接近，也就是说要尽量往正方形上靠，其次，对于n1,n2,n3都是有要求的，n1=n3<=n2,并且3<=n2<=N,而n1,n2,n3的定义分别是：n1,n3是这个U形的高，n2为这个U形的宽,因此在初步计算n1的时候需要用到n1=(N+2)/3,因为三条边一共有两个重合点，这两个重合点需要计算进去，然后要对当前的n1,n2,n3进行判断，看是否符合题目条件，不符合则进行调整，这一部分我使用了一个while循环，很简单就不赘述了，下面还是照例放上我的代码。
#include<iostream>
#include<string>
using namespace std;
int main()
{string str;cin >> str;int N = str.length();int n1, n2, n3;n1 = n3 =(N+2)/3;n2 = N - n1 - n3;while (true){if (n1+1 <= n2 && n2 >= 3 && n2 <= N){break;}else{n1--;n3--;n2 += 2;}}for (int i = 0; i < n1; i++){cout << str[i];for (int j = 0; j < n2-2; j++){cout << " ";}cout << str[N-i-1];cout << endl;}for (int i = 0; i < n2; i++){cout << str[n1+i];}system("pause");
}