HDU1219--AC Me

阳光穿透心脏的1/2处 2022-08-27 13:40 140阅读 0赞

# AC Me #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 11549    Accepted Submission(s): 5123**

Problem Description

Ignatius is doing his homework now. The teacher gives him some articles and asks him to tell how many times each letter appears.  
  
It's really easy, isn't it? So come on and AC ME.

Input

Each article consists of just one line, and all the letters are in lowercase. You just have to count the number of each letter, so do not pay attention to other characters. The length of article is at most 100000. Process to the end of file.  
  
Note: the problem has multi-cases, and you may use "while(gets(buf))\{...\}" to process to the end of file.

Output

For each article, you have to tell how many times each letter appears. The output format is like "X:N".   
  
Output a blank line after each test case. More details in sample output.

Sample Input

hello, this is my first acm contest!
    work hard for hdu acm.

Sample Output

a:1
    b:0
    c:2
    d:0
    e:2
    f:1
    g:0
    h:2
    i:3
    j:0
    k:0
    l:2
    m:2
    n:1
    o:2
    p:0
    q:0
    r:1
    s:4
    t:4
    u:0
    v:0
    w:0
    x:0
    y:1
    z:0
    
    a:2
    b:0
    c:1
    d:2
    e:0
    f:1
    g:0
    h:2
    i:0
    j:0
    k:1
    l:0
    m:1
    n:0
    o:2
    p:0
    q:0
    r:3
    s:0
    t:0
    u:1
    v:0
    w:1
    x:0
    y:0
    z:0

解析：感觉这是一道非常基础但是自我感觉非常适合入门的字符串的题目，我是在杭电ACM Steps中进阶的时候做的，感觉不错所以拿来博客，记录一下，与众人分享哈！

#include <iostream>
    #include <string>
    #include <string.h>
    #include <ctype.h>
    using std::endl;
    using std::cin;
    using std::cout;
    using std::string;
    int result[26];
    int main()
    {
    #ifdef LOCAL
    	freopen("input.txt" , "r" , stdin);
    #endif
    	string str;
    	while(getline(cin , str))
    	{
    		memset(result , 0 , sizeof result);
    		for(int i=0; i<str.length();++i)
    		{
    			if(isalpha(str[i]))
    			{
    				result[str[i]-'a']++;
    			}
    		}
    		for(int i=0;i<26;++i)
    		{
    			cout << (char)('a'+i) << ":" <<result[i] << endl;
    		}
    		cout << endl;
    	}
    	return 0;
    }