TOJ 1547. To and Fro【栈和队列】

1547. To and Fro

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Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 1267 Accepted Runs:953

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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 yh p k n ne l e a ir a h s ge c o n hs e m o tn 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

5toioynnkpheleaigshareconhtomesnlewx3ttyohhieneesiaabss0

Sample Output

theresnoplacelikehomeonasnowynightxthisistheeasyoneab

之前做过这个题，当时是用二维数组直接模拟的，然后想换成栈和队列试试，用手机敲了很久，然后用电脑调试了一会才过.........

感觉这道题用stl 去模拟真实是浪费了，还是安安稳稳的用数组吧......

//#include<bits/stdc++.h>#include<stdio.h>#include<string.h>#include<stack>#include<queue>using namespace std;queue< char> q[51];stack< char > s[51];char x[205];int main(){int n,len;//freopen("shuju.txt", "r", stdin);while (scanf("%d", &n),n){//getchar();memset(x,0,sizeof(x));scanf("%s",x);len=strlen(x);while (len%n){x[len++]='x';}int i=0,cnt;while (i<len){cnt=0;while (i<len&&cnt<n){q[i/2/n].push(x[i]);++cnt;++i;}cnt=0;while (i<len&&cnt<n){s[i/2/n].push(x[i]);++cnt;++i;}}while (!q[0].empty()){for(int j=0;j<len/n;++j){if(!q[j].empty())//按道理这句话不需要加，但是去掉就不对了.....{printf("%c", q[j].front());q[j].pop();}if(!s[j].empty()){printf("%c", s[j].top());s[j].pop();}}}printf("\n");}return 0;}