Timus 1404. Easy to Hack! 有一个密码问题

When Vito Maretti writes an important letter he encrypts it. His method is not very reliable but it’s enough to make any detective understand nothing in that letter. Sheriff doesn’t like such state
of affairs. He wants to hack the cipher of Vito Maretti and he promises to forget about all of your faults if you do that for him. Detectives will tell you what this cipher looks like.

Each word is enciphered separately. Assume an example that consists only of the small Latin letters.

At first step every letter is replaced with the corresponding number: a with 0, b with 1, c with 2,
..., z with 25.Then 5 is added to the first number, the first number is added to the second one, the second number – to the third one and so on. After that if some number exceeds 25 it is replaced with the residue of division
of this number by 26. And then those numbers are replaced back with the letters.

Let’s encipher the word secret.

Step 0.   s   e   c   r   e   t
Step 1.   18  4   2   17  4   19
Step 2.   23  27  29  46  50  69
Step 3.   23  1   3   20  24  17
Step 4.   x   b   d   u   y   r

We’ve got the word xbduyr.

Input

You are given an encrypted word of small Latin letters not longer than 100 characters.

Output

the original word.

Sample

xbduyr

secret

原来简单的加密也不是什么难事。做了那么多加密题，我也可以随手写个加密器了。

本题就是一个字符串的操作， 考的是字符串和整形之间的转换操作：

本题注意：第一个字母小于f的时候，转换为整数就会小于5了。

#include <string>
#include <vector>
#include <cmath>
#include <algorithm>
#include <iostream>
using namespace std;

void EasytoHack1404()
{
	string s;
	cin>>s;
	vector<int> tmp(s.size());
	for (int i = 0; i < s.size(); i++)
	{
		tmp[i] = s[i] - ‘a‘;
	}
	int t = tmp[0];
	tmp[0] -= 5;
	if (tmp[0] < 0) tmp[0] += 26;
	for (int i = 1; i < tmp.size(); i++)
	{
		int a = tmp[i];
		tmp[i] -= t;
		if (tmp[i] < 0) tmp[i] += 26;
		t = a;
	}
	for (int i = 0; i < tmp.size(); i++)
	{
		cout<<char(tmp[i]+‘a‘);
	}
}

Timus 1404. Easy to Hack! 有一个密码问题,码迷,mamicode.com