To and Fro
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4653 Accepted Submission(s): 3248
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.
letters. The last input set is followed by a line containing a single 0, indicating end of input.
5 toioynnkpheleaigshareconhtomesnlewx 3 ttyohhieneesiaabss 0
theresnoplacelikehomeonasnowynightx thisistheeasyoneab
解题思路：将一个字符串存入二维数组里（若行号是从零开始，则行号为奇数是正序，行号为偶数时逆序，详情见下图），
例如：输入toioynnkpheleaigshareconhtomesnlewx，输入到二维数组的顺序
<pre name="code" class="cpp">#include<cstdio>
#include<cstring>
int main()
{
int n,i,j,k,len,t;
char a[1010][1010],s[1010];
while(scanf("%d",&n)&&n)
{
scanf("%s",s);
len=strlen(s);
t=len/n; //t为行数
k=0;
for(i=1; i<=t; i++)
{
if(i%2==1)
{
for(j=1; j<=n; j++)
a[i][j]=s[k++];
}
else
{
for(j=n; j>=1; j--)
a[i][j]=s[k++];
}
}
for(j=1; j<=n; j++)
{
for(i=1; i<=len/n; i++)
{
printf("%c",a[i][j]);
}
}
printf("\n");
}
return 0;
}