【hdu 1219 AC Me】

AC Me

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7077 Accepted Submission(s): 3165

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

Author

Ignatius.L

Source

杭州电子科技大学第三届程序设计大赛

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Ignatius.L

 1 // Project name : 1219 ( AC Me ) 
 2 // File name    : main.cpp
 3 // Author       : Izumu
 4 // Date & Time  : Sun Jul  8 19:45:31 2012
 5 
 6 
 7 #include <iostream>
 8 using namespace std;
 9 
10 int main()
11 {
12     int count[26];
13     char ch;
14     while (cin.get(ch))
15     {
16         // init for count[26]
17         for (int i = 0; i < 26; i++)
18         {
19             count[i] = 0;
20         }
21         count[ch - 'a']++;
22 
23         while (cin.get(ch) && ch != '\n')
24         {
25             // cout << ch << endl;
26             if (ch >= 'a' && ch <= 'z')
27             {
28                 count[ch - 'a']++;
29             }
30         }
31 
32         // output 
33         for (int i = 0; i < 26; i++)
34         {
35             cout << (char) (i + 'a') << ":" << count[i] << endl;
36         }
37         cout << endl;
38     }
39     return 0;
40 }
41 
42 // end 
43 // ism