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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14100 Accepted Submission(s): 6194
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
竟然可以这么简单，不过我担心的是，定义的数组c只能存26,个数据，当遇到-65时会出现视情况，其实这个情况也没什么，超出了也没影响，超出时会给一个随机的不知道是什么东西的东西，反正你也不要输出
#include<stdio.h>
#include<string.h>
int main()
{
char a[100001], b[100001];
int c[26],s,i;
while(gets(a)!=NULL)
{
memset(c,0,sizeof(c));
for(i=0;a[i]!='\0';i++)
{
s=a[i]-'a';
c[s]++;
}
for(i=0;i<28;i++)
printf("%c:%d\n",'a'+i,c[i]);
printf("\n");
}
return 0;
}
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AC Me
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原文地址：http://blog.csdn.net/l15738519366/article/details/47323537
