首页 > ACM题库 > HDU-杭电 > HDU 4144-Bacon’s Cipher[解题报告]HOJ
2015
04-16

HDU 4144-Bacon’s Cipher[解题报告]HOJ

Bacon’s Cipher

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

35
ON1E2H5Q39AK2TGIC9ERT39B2P423L8B20D

样例输出:

FLEENOW

#include<iostream>
#include<cstdio>

using namespace std;

char ch[11111];

int main()
{
    int l, i, j, sum;
    while(cin >> l)
    {
        cin >> ch;
        for(i = 0; i < l; i += 5)
        {
            sum = 0;
            for(j = i; j < i + 5; j++)
            {
                sum *= 2;
                if(ch[j] >= '0' && ch[j] <= '9')
                    sum++;
            }
            cout << char(sum % 26 + 'A');
        }
        cout << endl;
    }
}

  1. 我没看懂题目
    2
    5 6 -1 5 4 -7
    7 0 6 -1 1 -6 7 -5
    我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
    第二个是7 0 6 -1 1 -6 7输出是14 7 7
    不知道题目例子是怎么得出来的