SRM 114 DIV2 [550]
来源:互联网 发布:mac设置用户头像 编辑:程序博客网 时间:2024/05/01 18:23
#include<vector>
#include<string>
using namespace std;
class BinaryCode
...{
public:
vector<string> decode(string message)
...{
vector<string> temp;
temp.push_back(message);
temp.push_back(message);
temp[0][0]='0';
temp[1][0]='1';
int len=message.size();
if (len==1)
switch (message[0])...{
case '0':
temp[1].assign("NONE");
break;
case '1':
temp[0].assign("NONE");
break;
default:
temp[0].assign("NONE");
temp[1].assign("NONE");
break;
}
else...{
for (int i=1;i!=len;++i)...{
if (i==1)...{
temp[0][i]=message[i-1]-temp[0][i-1]+'0';
temp[1][i]=message[i-1]-temp[1][i-1]+'0';
}
else...{
temp[0][i]=message[i-1]-temp[0][i-1]-temp[0][i-2]+'0'+'0';
temp[1][i]=message[i-1]-temp[1][i-1]-temp[1][i-2]+'0'+'0';
}
}
if (temp[0][len-1]+temp[0][len-2]!=message[len-1]+'0') temp[0].assign("NONE");
if (temp[1][len-1]+temp[1][len-2]!=message[len-1]+'0') temp[1].assign("NONE");
for (int i=1;i!=len;++i)...{
if (temp[0][i]>'1' || temp[0][i]<'0')...{
temp[0].assign("NONE");
}
if (temp[1][i]>'1' || temp[1][i]<'0')...{
temp[1].assign("NONE");
}
}
}
return temp;
}
};
Problem Statement
Let's say you have a binary string such as the following:
011100011
One way to encrypt this string is to add to each digit the sum of its adjacent digits. For example, the above string would become:
123210122
In particular, if P is the original string, and Q is the encrypted string, then Q[i] = P[i-1] + P[i] + P[i+1] for all digit positions i. Characters off the left and right edges of the string are treated as zeroes.
An encrypted string given to you in this format can be decoded as follows (using 123210122 as an example):
- Assume P[0] = 0.
- Because Q[0] = P[0] + P[1] = 0 + P[1] = 1, we know that P[1] = 1.
- Because Q[1] = P[0] + P[1] + P[2] = 0 + 1 + P[2] = 2, we know that P[2] = 1.
- Because Q[2] = P[1] + P[2] + P[3] = 1 + 1 + P[3] = 3, we know that P[3] = 1.
- Repeating these steps gives us P[4] = 0, P[5] = 0, P[6] = 0, P[7] = 1, and P[8] = 1.
- We check our work by noting that Q[8] = P[7] + P[8] = 1 + 1 = 2. Since this equation works out, we are finished, and we have recovered one possible original string.
Now we repeat the process, assuming the opposite about P[0]:
- Assume P[0] = 1.
- Because Q[0] = P[0] + P[1] = 1 + P[1] = 0, we know that P[1] = 0.
- Because Q[1] = P[0] + P[1] + P[2] = 1 + 0 + P[2] = 2, we know that P[2] = 1.
- Now note that Q[2] = P[1] + P[2] + P[3] = 0 + 1 + P[3] = 3, which leads us to the conclusion that P[3] = 2. However, this violates the fact that each character in the original string must be '0' or '1'. Therefore, there exists no such original string P where the first digit is '1'.
Note that this algorithm produces at most two decodings for any given encrypted string. There can never be more than one possible way to decode a string once the first binary digit is set.
Given a string message, containing the encrypted string, return a vector <string> with exactly two elements. The first element should contain the decrypted string assuming the first character is '0'; the second element should assume the first character is '1'. If one of the tests fails, return the string "NONE" in its place. For the above example, you should return {"011100011", "NONE"}.
Definition
Class: BinaryCode Method: decode Parameters: string Returns: vector <string> Method signature: vector <string> decode(string message) (be sure your method is public)Constraints
- message will contain between 1 and 50 characters, inclusive. - Each character in message will be either '0', '1', '2', or '3'.Examples
0)"123210122"
Returns: { "011100011", "NONE" }
The example from above.
1)"11"
Returns: { "01", "10" }
We know that one of the digits must be '1', and the other must be '0'. We return both cases.
2)"22111"
Returns: { "NONE", "11001" }
Since the first digit of the encrypted string is '2', the first two digits of the original string must be '1'. Our test fails when we try to assume that P[0] = 0.
3)"123210120"
Returns: { "NONE", "NONE" }
This is the same as the first example, but the rightmost digit has been changed to something inconsistent with the rest of the original string. No solutions are possible.
4)"3"
Returns: { "NONE", "NONE" }5)
"12221112222221112221111111112221111"
Returns: { "01101001101101001101001001001101001", "10110010110110010110010010010110010" }
This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.
- SRM 114 DIV2 [550]
- SRM 114 DIV2 [550]
- TopCoder SRM 144 DIV2(550-point)
- TopCoder SRM 144 DIV2 550points
- SRM 697 div2 550 推公式
- SRM 398 DIV2 [250]
- SRM 399 DIV2 [250]
- SRM 387 DIV2 [250]
- SRM 387 DIV2 [600]
- SRM 397 DIV2 [500]
- SRM 400 DIV2 [250]
- SRM 400 DIV2 [500]
- SRM 397 DIV2 [1000]
- SRM 405 DIV2
- SRM 421Div2 500
- SRM 457 Div2 500
- SRM 465(DIV1 DIV2)
- SRM 465(DIV1 DIV2)
- 时过境迁
- BeanShell快速入门---Java应用程序脚本引擎
- 真没想到,我都开始写博客了
- BeanShell桌面---Java应用程序脚本引擎
- Eclipse快捷键大全
- SRM 114 DIV2 [550]
- IBM Java Jvm GC实现内幕
- 对于C语言MFC的详细的解析(5) --深入MFC类库
- 网友与我关于一些GUI编程问题的对话实录
- C#编码规范
- 分析java + xml 的开发成功实例--webmail
- AJAX 无法解析 & 等字符 问题
- Log4j简明手册(一)
- 使用LogKit进行日志操作