Time limit: 1.0 second
Memory limit: 64 MB
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
Sample
| input |
output |
xbduyr
|
secret
|
Problem Author: Vladimir Yakovlev
Problem Source: The 12th High School Pupils Collegiate Programming Contest of the Sverdlovsk Region (October 15, 2005)
解析：直接按题意模拟即可。
AC代码：
#include <bits/stdc++.h>
using namespace std;
int a[102];
int main(){
string s;
while(cin>>s){
int n = s.size();
for(int i=0; i<n; i++) a[i] = s[i] - 'a';
if(a[0] < 5) a[0] += 26;
for(int i=1; i<n; i++){
while(a[i] < a[i-1]) a[i] += 26;
}
for(int i=n-1; i>0; i--) a[i] -= a[i-1];
a[0] -= 5;
for(int i=0; i<n; i++) printf("%c", a[i] + 'a');
puts("");
}
return 0;
}
URAL 1404. Easy to Hack! （模拟）
原文：http://blog.csdn.net/u013446688/article/details/46603901
