HDU 1200 To and Fro
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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
题目大意:
有多组测试数据,输入一个数n,当n=0时结束。接下来输入一个字符串,将字符串排列为n列。按照蛇形。然后将字符串列变行行变列输出。
c++
#include <iostream>#include<cstring>using namespace std;int main(){ char a[210]; char m[210][210]; int b,c,d,e,f,g; while(cin>>b) { if(b==0) break; cin>>a; f=strlen(a); g=f/b; c=0; while(1) { e=0; for(d=c*b;d<(c+1)*b;d++) { m[c][e]=a[d]; //将字符串存入一个二维字符数组中,因为这里我没用蛇形排列下面输出时有调整。 e++; } c++; if(c>=g) { break; } } for(e=0;e<b;e++) { for(d=0;d<c;d++) { if(d%2==0) //因为开始存时没有按照蛇形,数组又是从0开始,所以当碰到奇数时将其倒过来输出 { cout<<m[d][e]; } else { cout<<m[d][b-e-1]; } } } cout<<endl; } return 0;}
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