ACM HDU 1219 AC me（简单题，但是花了很长时间才AC）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1219

简单题。

收获：

gets()的使用。

AC Me

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5280    Accepted Submission(s): 2388

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

 

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int num[26];
char let[27]="abcdefghijklmnopqrstuvwxyz";
char str[20000];
int main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    int i;
    while(gets(str))
    {
        memset(num,0,sizeof(num));
        int len=strlen(str);
        //if(len==0)break;
        for(int i=0;i<len;i++)
        {
            if(str[i]>='a'&&str[i]<='z')
            {
                int t=str[i]-'a';
                num[t]++;
            }    
        } 
        for(int i=0;i<26;i++)
          printf("%c:%d\n",let[i],num[i]); 
        printf("\n");  
    }  
    return 0;  
}