题目：

AC Me

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13465    Accepted Submission(s): 5927

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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题目分析：

简单题。统计各个字母出现的次数。

代码如下：

/*
 * a.cpp
 *
 *  Created on: 2015年3月11日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>
#include <map>
#include <cstring>

using namespace std;

int main() {
	string str;

	int cnt[26];//开一个数组，用于存储每个字符出现的次数
	while (getline(cin, str)) {//读入一行数据.不要用C++提交,可能会CE
		memset(cnt, 0, sizeof(cnt));

		int len = str.length();
		int i;
		for (i = 0; i < len; ++i) {//遍历整个字符串
			if(str[i] >= ‘a‘ && str[i] <= ‘z‘){//如果该字符是一个字母
				cnt[str[i] - ‘a‘]++;//统计该字母出现的次数
			}
		}

		for(i = 0 ; i < 26 ; ++i){//遍历cnt数组
			printf("%c:%d\n",‘a‘+i,cnt[i]);//输出各个字母出现的次数
		}

		printf("\n");
	}

	return 0;
}