if(buff[0] == 'Q') ...Continue

#include <iostream.h>
#include <string.h>
#include <stdio.h>
void fill(char *buff1, int len);

int main(int argc, char* argv[])
{
      char buff1[50];
      char buff[50];
      char *searchString="a";
      int len=0, occurance=0;
      char *p;
      
        for(;;)
        {
               cout << "Enter Series of string. Q to quit" << endl;
                memset(buff, ' ', 50);
               cin >> buff;
              //len=strlen(strstr(buff, searchString));
              if( p=strstr(buff, searchString)){
                     len=strlen(p);
                     printf("%d\n",len);
              }else{
                     printf("(%s) not found in (%s)\n",searchString,buff);
                     len=0;
      cout << buff[0] << endl;  
      if(buff[0] == 'Q')  //Some Problem Here Does'nt exit, 2'nd time onwards
                return 0;
       else
                continue;
      }
             memset(buff1, ' ', strlen(buff1));
             strcpy(buff1, strstr(buff, searchString));
             fill(buff1, len);
             memset(buff, ' ', 50);
             while(len != 0)
             {
                      cout << "We are in line 31" << endl;
      occurance++;
      cout << occurance << endl << endl << endl;      
      cout << buff1 << endl;
      //len=strlen(strstr(buff1+1, searchString));
      if( p=strstr(buff1+1, searchString) ){
      len=strlen(p);
      printf("%d\n",len);
      }else{
             printf("(%s) not found in (%s)\n", searchString, buff1+1);
             len=0;
             break;
      }
                        
      if(len != 0)
      {
               strcpy(buff1, strstr(buff1+1, searchString));
               fill(buff1, len);
      }

      cout << "We are in line 43" << endl;
      //memset(buff1, ' ', len);
             }

       }
      
return 0;
}
void fill(char *buff1, int len)
{
         *(buff1+len)='\0';
}
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      if(buff[0] == 'Q')  //Some Problem Here Does'nt exit, 2'nd time onwards
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Please run this code and let me know if you have a fix.