#include
#include
unsigned?char?fn_feedback（int?n//級數為n，假若為5
						??unsigned?char?c//參數C1C2C3C4C5組合成c
						??unsigned?char?curr_state）//初始狀態
{
	unsigned?char?t?=?c?&?curr_state;//實現了c1&k1c2&k2c3&k3c4&k4c5&k5
	unsigned?char?s?=?t?&?（unsigned?char）128;//get?first?bit??s實現的是cj&kj的異或??即反饋多項式??
												//	（unsigned?char）128=“10000000“進行位于，就能夠分離出cj&kj了		
	for（int?i?=1;?i	{
		s?=?s?^?（?（t?<	}
	return?（unsigned?char）（（curr_state?<>?7））;?//return?next?state接下來的狀態便是上一狀態整體向右位移1位，
											//在或上s向右位移7位，因為s的形式是“10000000或者00000000”
}
void?lfsr_output_byte（int?n
					??unsigned?char?c
					??unsigned?char?init_state
					??unsigned?char?output_bytes[]
					??int?byte_length）//多

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????目錄???????????0??2013-12-08?11:15??stream_Cipher\
?????目錄???????????0??2013-12-08?11:10??stream_Cipher\Debug\
?????文件????????8643??2013-12-08?11:10??stream_Cipher\Debug\stream.obj
?????文件????????9456??2013-12-08?11:10??stream_Cipher\Debug\stream_Cipher.lib
?????文件??????187232??2013-12-08?11:02??stream_Cipher\Debug\stream_Cipher.pch
?????文件???????41984??2013-12-08?11:10??stream_Cipher\Debug\vc60.idb
?????文件???????61440??2013-12-08?11:10??stream_Cipher\Debug\vc60.pdb
?????文件????????1732??2013-12-08?11:10??stream_Cipher\stream.cpp
?????文件?????????458??2013-12-08?11:02??stream_Cipher\stream.h
?????文件????????1259??2013-12-08?11:08??stream_Cipher\stream_C.cpp
?????文件????????2966??2013-12-08?11:15??stream_Cipher\stream_Cipher.dsp
?????文件?????????534??2013-12-06?10:06??stream_Cipher\stream_Cipher.dsw
?????文件???????41984??2013-12-08?11:15??stream_Cipher\stream_Cipher.ncb
?????文件???????48640??2013-12-08?11:15??stream_Cipher\stream_Cipher.opt
?????文件?????????800??2013-12-08?11:10??stream_Cipher\stream_Cipher.plg