HDU 1.2.7 AC Me

计数字母出现次数

最新推荐文章于 2017-12-06 09:23:28 发布

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本文介绍了一个简单的程序设计问题：统计一篇文章中每个英文字母出现的次数。文章提供了完整的C语言代码实现，包括如何读取输入、过滤非字母字符及最终输出统计结果。

AC Me

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

Total Submission(s): 4673 Accepted Submission(s): 1632

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

#include <stdio.h>  
#include <string.h>  
#include <ctype.h>  
  
int main()  
{  
int i, len;  
char s[100000];  
while (gets(s))  
{  
int a[26] = { 0 };  
len = strlen(s);  
for (i = 0; i<len; i++)  
{  
if (isalpha(s[i]))  
a[s[i] - 'a']++;  
}  
for (i = 0; i<26; i++)  
{  
printf("%c:%d\n", i + 'a', a[i]);  
}  
printf("\n");  
}  
return 0;  
}