To and Fro（字符串规律）

最新推荐文章于 2021-03-02 15:34:19 发布

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Mo和Larry设计了一种独特的加密方式，通过设定列数并填充额外的随机字母形成矩形矩阵来加密消息。接收方Larry需将加密后的行按照交替左右的方式还原成原始消息。本文提供了解码算法并附上示例输入输出。

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To and Fro

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 13 Accepted Submission(s) : 9

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Problem Description

Mo and Larry have devised a way of encrypting messages. They first decide secretly on the number of columns and write the message (letters only) down the columns, padding with extra random letters so as to make a rectangular array of letters. For example, if the message is “There’s no place like home on a snowy night” and there are five columns, Mo would write down

t o i o y
h p k n n
e l e a i
r a h s g
e c o n h
s e m o t
n l e w x

Note that Mo includes only letters and writes them all in lower case. In this example, Mo used the character ‘x’ to pad the message out to make a rectangle, although he could have used any letter.

Mo then sends the message to Larry by writing the letters in each row, alternating left-to-right and right-to-left. So, the above would be encrypted as

toioynnkpheleaigshareconhtomesnlewx

Your job is to recover for Larry the original message (along with any extra padding letters) from the encrypted one.

Input

There will be multiple input sets. Input for each set will consist of two lines. The first line will contain an integer in the range 2. . . 20 indicating the number of columns used. The next line is a string of up to 200 lower case letters. The last input set is followed by a line containing a single 0, indicating end of input.

Output

Each input set should generate one line of output, giving the original plaintext message, with no spaces.

Sample Input

5
toioynnkpheleaigshareconhtomesnlewx
3
ttyohhieneesiaabss
0

Sample Output

theresnoplacelikehomeonasnowynightx
thisistheeasyoneab


其实可以发现一个规律，创建一个二维数组，奇数行时从左到右赋值，偶数行时从右到左
ac代码：
#include<stdio.h>
#include<string.h>
int main()
{
	char s[205];
	char s1[20][20];
	int h,j,i;
	int n;
	while(scanf("%d",&n)!=EOF&&n)
	{
		h=1;j=1;
		scanf("%s",s);
		int len=strlen(s);
		for(i=0;i<len;i++)
		{
			if(h%2==1)//奇数行 
			{
				if(j<=n)
			    s1[h][j++]=s[i];
			    else
			    {
			    	i--;h++;j--;
			    }
			}
			else//偶数行 
			{
				if(j>=1)
				s1[h][j--]=s[i];
				else
				{
					i--;h++;j++;
				}
			}
		}
		for(j=1;j<=n;j++)
		for(i=1;i<=h;i++)
		{
			printf("%c",s1[i][j]);
		}
		printf("\n");
	}
	return 0;
}