群聯
筆試給了試題和兩張白紙，第一題就是寫出keygen，
以前也有寫過JS版本的
分析:
面試主要是Linux相關所以用C/C++撰寫
C的版本在於只有兩張紙還有大約10題，必定不適合寫個Link list...
於是直接alloc裝Key...
1. 寫一個keygen 0-F 6個字gen出不重複的key
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#define ARRY_SIZE(A) (sizeof(A)/sizeof(A[0]))
char allowChar [] =
{
'0','1','2','3','4','6','7','8','9','A','B','C','D','E','F'
};
char *keyList;
char getChar(){
return allowChar[rand()%ARRY_SIZE(allowChar)];
}
bool inList( char * keyListPtr , int currentListLength , int keySize ,char *tmpKey )
{
char *keyPtr;
int i;
for(i = 0 ; i < currentListLength; i ++ )
{
keyPtr = keyListPtr+ i*keySize;
if( strcmp(keyPtr ,tmpKey ) == 0)
return true;
}
return false;
}
char *genKey( char *keyListPtr ,int keyNumber,int Length){
int i,j,totalBufferSize,keySize;
keySize = Length +1 ;
totalBufferSize = keyNumber*keySize;
char *tmpKey = (char *)malloc(keySize);
keyListPtr = (char * ) malloc( totalBufferSize );
memset(keyListPtr , 0 , totalBufferSize);
j = 0;
while( j < keyNumber )
{
memset(tmpKey, 0 ,keySize);
for( i = 0 ; i < Length ; i++ )
*(tmpKey+i) = getChar();
if( !inList( keyListPtr , j , keySize ,tmpKey ) )
{
memcpy( keyListPtr + j*keySize , tmpKey , keySize);
j++;
}
}
free(tmpKey);
printf("Gen Done!\n");
return keyListPtr;
}
void display( char * keyListPtr , int currentListLength , int keySize ){
char *keyPtr;
int i;
for(i = 0 ; i < currentListLength; i ++ )
{
keyPtr = keyListPtr + i*keySize;
printf("key[%4d] : %s\n",i,keyPtr);
}
return ;
}
int main()
{
int keyNumber = 30;
int keyLength = 6;
char * keyList;
keyList = genKey( keyList , keyNumber , keyLength );
display(keyList ,keyNumber , keyLength + 1 );
free(keyList);
return 0;
}
find gcc in linux kernel
find /usr/ -type f -name "gcc"
