题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1200
To and Fro
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5129 Accepted Submission(s): 3550
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
题目大意:输入按照题目中列举的矩阵输入，假使从第0行开始，也就是偶数行按序输入，奇数行反序输入。最后输出的时候是按列输出~
有两种方法：第一种是DP，用dp这个数组来按照原有矩阵进行存放。最后直接输出dp即可。第二种就是直接模拟，比如第一列的就是，0,9,10,19,20,29,30；将其分成两组，第一组由0,10,20,30组成，第二组由9,19,29组成；进一步解释就是第一组的（位置数-i）%n==0；第二组的（位置数+i+1）%n==0；
详见代码。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4
5 using namespace std;
6
7 char ch[210];
8 char dp[110][20];
9
10 int main ()
11 {
12 int n;
13 while (~scanf("%d",&n),n)
14 {
15 scanf("%s",ch);
16 int x=0,y=0;
17 int len=strlen(ch);
18 for (int i=0; i<len; i++)
19 {
20 dp[x][y]=ch[i];
21 if (x%2==0)
22 {
23 y++;
24 if (y==n)
25 {
26 x++;
27 y--;
28 }
29 }
30 else
31 {
32 y--;
33 if (y==-1)
34 {
35 x++;
36 y++;
37 }
38 }
39
40 }
41 for (int i=0; i<n; i++)
42 {
43 for (int j=0; j<x; j++)
44 {
45 printf ("%c",dp[j][i]);
46 }
47 }
48 printf ("\n");
49 }
50 return 0;
51 }
第二种模拟代码，详见注释
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4
5 using namespace std;
6
7 int main()
8 {
9 int n;
10 char ch[250];
11 while(scanf("%d",&n),n)
12 {
13 scanf("%s",ch);
14 int l=strlen(ch);
15 for(int i=0;i<n;i++) //i表示第几列
16 {
17 for(int j=0;j<l;j++) //j表示第几个数
18 {
19 if((j/n)%2==0) //判断奇偶行
20 {
21 if((j-i)%n==0) //满足第一组的条件
22 printf("%c",ch[j]);
23 }
24 else
25 {
26 if((j+i+1)%n==0) //满足第二组的条件
27 printf("%c",ch[j]);
28 }
29 }
30 }
31 printf("\n");
32 }
33 return 0;
34 }