What's The Frequency, Kenneth? 

 

#include 
    
     main(){  int i;  char *suffix[]= { "st", "nd", "rd" };  char *item[]= { "Unix" , "cat", "sed", "awk", "grep", "ed", "vi"};   printf("In the beginning, there was nothing.\n");  for (i= 0; i < 7; i++)    printf("And on the %d%s day, God created %s. And it was good.\n",           i + 1, (i < 3) ? suffix[i] : "th", item[i]);}
    

But then God saw that vi led people into temptation. Instead of choosing the righteous ways of make, dbx, andRCS, people used long command lines, printf(), and tape backups.

So God decreed, ``I see that Engineers have thus defiled my vi. And so, I shall create emacs, an editor more powerful than words. Further, for each instantiation vi hitherto, the Engineer responsible shalt perform Penance. And lo, the Penance wilt be painful; there will be much wailing and gnushingof teeth. The Engineer will read many lines of text. For each line of text, the Engineer must tell me which letters occur the most frequently.''

``I charge you all with My Golden Rule: 'Friends shalt not let friends use vi'.''

 

Input and Output

Each line of output should contain a list of letters that all occured with the highest frequency in the corresponding input line, followed by the frequency.

The list of letters should be an alphabetical list of upper case letters followed by an alphabetical list of lower case letters.

 

Sample Input

 

When riding your bicycle backwards down a one-way street, if thewheel falls of a canoe, how many ball bearings does it take to fillup a water buffalo?Hello Howard.

 

Sample Output

 

e 6al 7a 3Hlo 2 学会使用memset函数

void *memset(void *s,int ch,size_t n);

函数解释：将 s 中前 n 个字节用 ch 替换并返回 s 。

memset：作用是在一段内存块中填充某个给定的值，它是对较大的结构体或数组进行清零操作的一种最快方法。

 

如把一个char a[20]清零，一定是 memset(a,0,20);

(地址，要赋予的值，连续几个字节)

 

#include 
     
      #include 
      
       #define MAX(x,y) (x>y?x:y)char str[1234567];int num[256];int main(){    int i;    int max;    while(gets(str))    {        if(str[0]=='\0')continue;        memset(num,0,sizeof(num));        for(i=0;str[i]!='\0';++i)++num[(int)str[i]];        max = 0;        for(i='A';i<='Z';++i)max = MAX(max,num[i]);        for(i='a';i<='z';++i)max = MAX(max,num[i]);        for(i='A';i<='Z';++i)if(num[i]==max)putchar(i);        for(i='a';i<='z';++i)if(num[i]==max)putchar(i);        printf(" %d\n",max);    }    return 0;}