九度 题目1464:Hello World for U

本文介绍了一种将任意长度的字符串以U形排列输出的方法。该算法通过计算字符串的长度来确定U形的各部分长度,并按特定顺序输出字符。适用于处理长度大于等于5且不超过80的字符串。

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#include <iostream>
#include <string.h>
#include <queue>
#include <map>
using namespace std;

int main()
{
    char str[81];
    while(cin>>str)
    {
        int n=strlen(str);
        //cout<<n<<endl;
        int n1=(n+2)/3;
        n1--;
        int n2=n-2*n1;
        for(int i=0;i<n1;i++)
        {
            cout<<str[i];
            for(int j=1;j<n2-1;j++)
                cout<<" ";
            cout<<str[n-i-1]<<endl;
        }
        for(int i=n1;i<n1+n2;i++)
            cout<<str[i];
        cout<<endl;
    }
    return 0;
}
算法设计思想:输入一个字符串输出U的形状,分三部分计算,n1是U的左半侧,n2是U的下半侧,n3与n1对称,根据所给的n1+n2+n3=n+2,(其中n是字符串的长度),然后注意空格循环打印出来就好,for循环错开。
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.
来源:
2012年浙江大学计算机及软件工程研究生机试真题
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