HDU 1328 IBM Minus One

IBM Minus One

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5354    Accepted Submission(s): 2742

Problem Description

You may have heard of the book '2001 - A Space Odyssey' by Arthur C. Clarke, or the film of the same name by Stanley Kubrick. In it a spaceship is sent from Earth to Saturn. The crew is put into stasis for the long flight, only two men are awake, and the ship is controlled by the intelligent computer HAL. But during the flight HAL is acting more and more strangely, and even starts to kill the crew on board. We don't tell you how the story ends, in case you want to read the book for yourself :-)

After the movie was released and became very popular, there was some discussion as to what the name 'HAL' actually meant. Some thought that it might be an abbreviation for 'Heuristic ALgorithm'. But the most popular explanation is the following: if you replace every letter in the word HAL by its successor in the alphabet, you get ... IBM.

Perhaps there are even more acronyms related in this strange way! You are to write a program that may help to find this out.

 

Input

The input starts with the integer n on a line by itself - this is the number of strings to follow. The following n lines each contain one string of at most 50 upper-case letters.

 

Output

For each string in the input, first output the number of the string, as shown in the sample output. The print the string start is derived from the input string by replacing every time by the following letter in the alphabet, and replacing 'Z' by 'A'.

Print a blank line after each test case.

 

Sample Input

2HALSWERC

 

Sample Output

String #1IBMString #2TXFSD

 

题目大意：

题目说了一大堆如题词（也就是废话），我们不管他恩，看重点，就是你只要把‘HAL’的每个字符向前面移动一位，你就能够找到'IBM'了，那么解密工作就变得很简单了，但是会有一个问题 那么出现'Z'的时候我们拿什么替换他，找到真正的密文呢？

辛好题目中有给： ‘Z----->>'A'   

分析：

我们可以从ASCLL编码的角度去考虑恩，只要将现在的字符+1就能得到相应的密文了  如果遇到了'Z'那么就用'A';代替就好了，其实我们可以发现这个密文解密是一个圈圈的；

脑补下：

 A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y  Z

 B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y  Z  A

给出AC代码：

#include<iostream>#include<string>#include<map>using namespace std;int main(){string str;int n, count = 0;cin >> n;getchar();while (n--){cin >> str;count++;int len = str.length();cout << "String #" << count << endl;for (int i = 0; i < len; i++){if (str[i] != 'Z')printf("%c", str[i] + 1);else printf("%c", 'A');}printf("\n\n");}return 0;}