TJU 2468 Counting Letters

　　　　　　　　　　　　　　　　　　　　　　2468 Counting Letters 

As a talented student, your boss gave you a task. Given a text string, you should find out which letters appear most frequently.

Really simple, isn't it?

Input

The first line of the input is the number of test cases. Then some test cases followed.

Each test case contains only one line, indicating the string mentioned above.

You can assume there is only lower case letters in the string. The length of the string is more than 0 and no more than 100.

Output

You should output one line for each test case, indicating the most frequent letters. Please note there may be more than one such letter, so you should output all of them, and these letters should be in the same order as the alphabet, that is, 'a', 'b', 'c', 'd' ... 'z'.

Sample Input

2
abcda
tjucstju

Sample Output

a
jtu

Hint

In the first sample test case, 'a' appears twice, while 'b','c','d' only appear once. So 'a' is the most frequent letter.

In the second sample test case, 't', 'j' and 'u' appear twice, so you should output them all, and 'j' should be the first.

Please note you should not output any unnecessary spaces or empty lines, or else you may get 'Presentation Error'.

解题报告：好久没有在OJ上做题了，今天有人问我这道题怎么做，嘿嘿，就写了写，题意就是统计字母出现最多的，如果有多个字母出现一样就按字母的前后顺序输出，

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX = 110;
int main()
{
    int n, i, max, len, num;
    char c[MAX];
    int count[30];
    scanf("%d", &n);
    while (n --)
    {
        memset(count, 0, sizeof(count));
        scanf("%s", c);
        len = strlen(c);
        for(i = 0; i < len; i ++)
        {
            num = c[i] - 'a';
            count[num]++;
        }
        max = 0;
        for (i = 0; i < 26; i ++)
        {
            if(max < count[i])
            {
                max = count[i];
            }
        }
        for (i = 0; i < 26; i ++)
        {
            if (max == count[i])
            {
                printf("%c", i + 'a');
            }
        }
        printf("\n");
    }
    return 0;
}