HDU 1219 AC Me

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本题为ACM竞赛题目，要求统计输入文本中每个小写字母的出现次数，并按特定格式输出。提供了完整的C语言代码实现，使用gets()读取输入并遍历字符串进行计数。

AC Me

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7582 Accepted Submission(s): 3410

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

Recommend

Ignatius.L

分析：检查字符串里各个字母出现的次数。。

代码：

#include<stdio.h>
#include<string.h>

char str[100010];
int a[26];

int main()
{
    while(gets(str))
    {
        int len,i;
        char ch='a';
        memset(a,0,sizeof(a));
        len=strlen(str);
        for(i=0;i<len;i++)
        {
            if(str[i]>='a'&&str[i]<='z')
                a[str[i]-'a']++;
        }
        for(i=0;i<26;i++)
        {
            printf("%c:",ch+i);
            printf("%d\n",a[i]);
        }
        printf("\n");
    }
    return 0;
}