PTA练习——Hello World for U
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
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.
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!
样例输出
h ! e d l l lowor
思路：
n1和n3是左右两条竖线从上到下的字符个数，n2是底部横线从左到右的字符个数。
题目中给了公式n1 + n2 + n3 – 2 = N，并且我们可以知道n1=n3,n1<=n2,在满足上述条件下n1要最大。
令N=字符串长度+2，进行分类讨论：
- 如果N % 3 == 0，n正好被3整除，直接n1 == n2 == n3;
- 如果N % 3 == 1，因为n2要比n1大，所以把多出来的那1个给n2
- 如果N % 3 == 2, 就把多出来的那2个给n2
所以得到公式：n1 = N / 3，n2 = N / 3 + N % 3
即步骤为：
1.计算字符串长度l；
2.计算两边的字符数n1= N / 3；
3.计算最后一行中间的字符数（前面每行中间的空格数）space=n2-2；
4.输出每行相应的字符。
#include <iostream>
using namespace std;
int main()
{
int i,j,n1,l,cnt=0,n2;
string s;
while(cin>>s){
l=s.length();
n1=(l+2)/3;
n2=n1+(l+2)%3;
for(i=1;i<n1;i++){
cout<<s[cnt];
for(j=1;j<=n2-2;j++){
cout<<" ";
}
cout<<s[l-cnt-1]<<endl;
cnt++;
}
for(i=cnt;i<cnt+n2;i++){
cout<<s[i];
}
}
return 0;
}
