HDU4144 Bacon's Cipher

#include<stdio.h>                                        #include<math.h>                      int main()                            {                                      int n,i,c,k,a[10010];                 char s[10010],ch[10010];                                                    while(~scanf("%d",&n)) { scanf("%s",s);                                                                                     k=(n/5)-1;                                                                  for(i=0;i<n;i++)                      {                                     if(s[i]>=65&&s[i]<=90)s[i]='0';       else if(s[i]>=48&&s[i]<=57)s[i]='1';  }                                                                           int m=0;                              int sum=0;                            for(i=n-1;i>=0;i--)                   {                                     sum=sum+((s[i]-48)*pow(2,m));         m++;                                  if(m==5){m=0;a[k]=sum;k--;sum=0;}     }                                     k=(n/5)-1;                            for(i=0;i<=k;i++)                     ch[i]='A'+a[i]; ch[k+1]='\0';                                                                                           printf("%s\n",ch);                      } return 0;                            }                                     

Bacon's Cipher

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1321 Accepted Submission(s): 727

Problem Description

Bacon's cipher or the Baconian cipher is a method of steganography (a method of hiding a secret message as opposed to a true cipher) devised by Francis Bacon. A message is concealed in the presentation of text, rather than its content.
As we all know, each letter has its position in the alphabet, ‘A’ is 0, ‘B’ is 1, ‘C’ is 2…and so on. And each number can be represented in binary code, for example, 2 is ‘10’ in binary system. Then we expand the binary code to five digits by adding leading zeros, then 10 becomes 00010. Now we can use this number to encode. To simplify the question, we define the rules as below:
0 corresponds to a random uppercase letter and 1 corresponds to a random number, so after encoding, 00010 ( ‘C’ ) is transformed to ABC1D or JUG9N.
To decode, do the opposite way around.

Input

The first line contains a positive number l, represents the length of the encoded string. L<=10000 and can be divided by 5. The second line is the encoded string.

Output

The original string.

Sample Input

35ON1E2H5Q39AK2TGIC9ERT39B2P423L8B20D

Sample Output

FLEENOW

Author

BNU