HDU1219 AC Me【输入输出】

AC Me

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

Total Submission(s): 18568    Accepted Submission(s): 7850

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

问题链接：HDU1219 AC Me。

题意简述：参见上文。

问题分析：这是一个字符统计题，只统计小写字母。

程序说明：

这里给出两个C语言程序，一个没有使用数组（正解），另外一个是使用了数组来存储字符串。

由于函数gets()不被推荐使用（容易造成存储越界访问），所有使用函数fgets()来读入一行字符串。

使用函数fgets()时，需要考虑字符串末尾的'\n'和'\0'，关系大数组的大小。

需要注意行结束条件！

题记：存储能省则省，只要不影响程序的简洁性。

AC的C语言程序如下：

/* HDU1219 AC Me */

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

#define N 26
int count[N];

int main(void)
{
    char c;

    while((c = getchar()) != EOF) {
        memset(count, 0, sizeof(count));

        while(c != EOF && c != '\n') {
            if(islower(c))
                count[c - 'a']++;

            c = getchar();
        }

        int i;
        for(i=0; i<N; i++)
            printf("%c:%d\n", 'a' + i, count[i]);
        printf("\n");
    }
    return 0;
}

AC的C语言程序如下：

/* HDU1219 AC Me */

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

#define N 100000 + 10
#define N2 26
char s[N+1];
int count[N2];

int main(void)
{
    int i;

    while(fgets(s, N, stdin) != NULL) {
        memset(count, 0, sizeof(count));

        i = 0;
        while(s[i] != '\n') {
            if(islower(s[i]))
                count[s[i] - 'a']++;
            i++;
        }

        for(i=0; i<N2; i++)
            printf("%c:%d\n", 'a' + i, count[i]);
        printf("\n");
    }

    return 0;
}