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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  de  ll  rlowo

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   dl   llowor

题意：

给定N（≥5）个字符的任何字符串，系统会要求您将字符形成U形。例如，helloworld可以打印为：

h  de  ll  rlowo

也就是说，必须按照原始顺序打印字符，从左上垂直线开始以n‐1个字符开始打印，然后沿着底部行以n‐2个字符从左到右开始打印，最后以-沿垂直线向上带有n个3个字符。而且，我们希望U尽可能平方-也就是说，必须满足n 1 = n 3 = max {k |对于所有3≤n2≤N}，k≤n2（n≥1 2 + n 2 2 + n 3 −2 = N.

输入规格：

每个输入文件包含一个测试用例。每个案例包含一个字符串，一行中不少于5个字符且不超过80个字符。该字符串不包含空格。

输出规格：

对于每个测试用例，按照描述中的指定，以U形打印输入字符串。

样本输入：

helloworld!

样本输出：

h   !e   dl   llowor

思路：

先确定n1,n2,n3的长度，然后模拟一遍就行了

代码：

#include <iostream>#include <algorithm>#include <vector>#include <cstdio>#include <cstring>#include <cstdio>#include <cmath>#include <queue>#include <set>#include <map>using namespace std;  int main() {     string s;  getline(cin,s);  int len=s.length(); //总长度  char site[40][40];  //地图  int n1,n2,n3;  n1 = n3 = (len+2)/3;   //因为n1<=n2,n1=n3,所以直接平分取整，多出来的给n2  n2 = len+2-n1-n3;    while(n2<3) n2+=2,n1--,n3--;  /*题目还规定n2要>=3，所以还要判断一下，不够我发现貌似不加也可以过  ，不知道是我理解错了题目，还是这数据没那么严格*/    for(int i=0;i<n1;i++)  //把地图先全部更新为空格    for(int j=0;j<n2;j++) //地图大小 n1行 n2列      site[i][j]=' ';        int c=0; //遍历字符串的位置  for(int i=0;i<n1;i++) //模拟第一列  site[i][0]=s[c++];       for(int j=1;j<n2-1;j++) //模拟最后一行  site[n1-1][j]=s[c++];    for(int i=n1-1;i>=0;i--) //模拟最后一列  site[i][n2-1]=s[c++];    for(int i=0;i<n1;i++)  //输出即可  {     	for(int j=0;j<n2;j++)      cout<<site[i][j];     cout<<endl;  }  return 0;}