#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。
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题目描述:
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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 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.
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输入:
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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.
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输出:
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For each test case, print the input string in the shape of U as specified in the description.
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样例输入:
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helloworld! ac.jobdu.com
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样例输出:
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h ! e d l l lowor a m c o . c jobdu.