2013年4月_武科大程序设计大赛_解题报告(problem 1002)

To and Fro

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3384 Accepted Submission(s): 2393



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
字符串的处理首先输入一个数n，将整个字符串按照“S形”的排列方式写出，每行n个字符。然后按照竖排的顺序依次输出。可以采用直接模拟的方式把S形数组转化为从左向右，从上到下的一维数组，然后每隔n个数输出一个字符。
#include<string.h>
#include<stdio.h>
int main()
{
	char a[10000];
	int n;
	char b[10000];
	while(scanf("%d",&n),n)
	{
		getchar();
		gets(a);
		int i=0,k;
		int length=strlen(a);
		int top=0;
		for(i=0;i<=((length-1)/n);i++)		//把S形存入b数组中
		{
			if(i%2==0)
			{
				for(k=i*n;k<=i*n+n-1;k++)
					b[top++]=a[k];
			}
			else
			{
				for(k=i*n+n-1;k>=i*n;k--)	
				{
				//	if(!a[k]) continue;
					b[top++]=a[k];
			
			}		}
		}
		b[top]=0;
		for(k=0;k<n;k++)				//每隔n-1个输出一个字符 
		for(int j=k;j<length;j+=n)
		{
		//	if(b[j])
			putchar(b[j]);
		}
		putchar(10);
	}
		
	return 0;
}