hdu 1219（1.2.8）

2024-06-13 04:58

文章标签 hdu 1.2 1219

本文主要是介绍hdu 1219（1.2.8），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目描述：AC Me

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

Total Submission(s): 3515 Accepted Submission(s): 1235

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:0a: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

#include<stdio.h>#include<stdlib.h>#include<string.h>int findout(char a){if(a>='a'&&a<='z')return a-'a'+1;elsereturn 0;}int main(){int len,i,c[27];char s[100005];while(gets(s)){memset(c,0,sizeof(c));len=strlen(s);for(i=0;i<len;i++)if(findout(s[i]))c[findout(s[i])]++;for(i=1;i<=26;i++)printf("%c:%d\n",'a'+i-1,c[i]); printf("\n");memset(s,0,sizeof(s));}return 0;}

总结：1.gest()读取一行的字符串，接受空格
2.返回s的长度，不包括结束符NULL。
3.不要轻易用while(len--)

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