(HDUStep 1.2.7)IBM Minus One(字符串运算)

题目：

      

IBM Minus One

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3943 Accepted Submission(s): 1519

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 2 HAL SWERC

Sample Output String #1 IBM String #2 TXFSD

Source Southwestern Europe 1997, Practice

Recommend Ignatius.L

题目分析：

      简单的字符串运算。例如： 'B' = 'A'+1 .....将给出的字符串中的每个字符加1后输出即可。

代码如下：

/*
 * h.cpp
 *
 *  Created on: 2015年1月28日
 *      Author: Administrator
 */
#include <iostream>
#include <cstdio>

using namespace std;


int main(){
	int n;
	scanf("%d",&n);

	int i;
	for(i = 0 ; i < n ; ++i){
		string str;
		cin >> str;

		int length = str.length();
		int j;
		for(j = 0 ; j < length ; ++j){
			if(str[j] == 'Z'){
				str[j] = 'A';
			}else{
				str[j] += 1;
			}
		}

		cout << "String #" << (i+1) <<endl;
		cout<<str<<endl<<endl;
	}

	return 0;
}