PAT甲级A1031 Hello World for U (20)

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

题意：给定一个字符串，让你打印成‘u'型。 要求如下：1.n1==n3。 2.n2>=n1。 3.n1+n2+n3=N+2。

思路：依题意有2*n1+n2=N+2===>n2=N+2-2*n1.

           并且n2>=n1可得:

           n1<=(N+2)/3;

           n2=N+2-2*n1;

将字符按要求存储在二维数组中然后打印即可，注意二维数组一定要初始化为空格！！！！否则会出错。

参考代码：

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
using namespace std;
char ans[80][80];
int main()
{
	int n,n1,n2,k=0;
	memset(ans,' ',sizeof(ans));
	string s;
	cin>>s;
	n=s.size();
	n1=(n+2)/3;
	n2=n+2-2*n1;
	for(int i=0;i<n1-1;i++) ans[i][0]=s[k++];
	for(int j=0;j<n2;j++) ans[n1-1][j]=s[k++];
	for(int i=n1-2;i>=0;i--) ans[i][n2-1]=s[k++];
	for(int i=0;i<n1;i++){
		for(int j=0;j<n2;j++){
			printf("%c",ans[i][j]);
		}
		printf("\n");
	}
	return 0;
}