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            <P align=center><B>IBM的MARS加密算法实现(下)</B><BR>作者:<A 
            href="mailto:hitwz@163.com">西安 吴真</A></P>
            <P><A 
            href="http://www.vckbase.com/document/viewdoc.asp?id=401">IBM的MARS加密算法实现(上)</A><BR><B><BR>2.3 
            密文解密</B><BR>用于密文解密的40个子密钥的生成和明文加密时的40个子密钥的生成方法相同.<BR><BR><B>2.3.1 
            第一步前向混合</B><BR>输入的128位密文分成四块D[0],D[1],D[2],D[3],选取生成的40个密钥的最后四个分别与上述四块数据进行加操作,<BR>D[0] 
            += K[36];<BR>D[1] += K[37];<BR>D[2] += K[38];<BR>D[3] += 
            K[39];<BR>结果作为第一轮操作的输入数据.<BR>第一轮:<BR></P>
            <TABLE cellSpacing=1 cellPadding=0 width=400 bgColor=#000000 
            border=0>
              <TBODY>
              <TR align=middle bgColor=#ffffff>
                <TD colSpan=4>D[0]</TD>
                <TD width=81>D[1]</TD>
                <TD width=110>D[2]</TD>
                <TD width=67>D[3]</TD></TR>
              <TR align=middle bgColor=#ffffff>
                <TD width=30>b0</TD>
                <TD width=34>b1</TD>
                <TD width=37>b2</TD>
                <TD width=41>b3</TD>
                <TD width=81>FirstTarget</TD>
                <TD width=110>SecondTarget</TD>
                <TD 
            width=67>ThirdTarget</TD></TR></TBODY></TABLE><BR>把D[0],D[1],D[2],D[3]合并成128位的数据,循环左移32位后分成四块D[0],D[1],D[2],D[3]其中D[0]作为源数据(Source),剩下的3个作为目标数据,把D[0]循环右移24位后的结果返回给D[0]<BR>把32位的源数据D[0]分成8位的四块b0,b1,b2,b3<BR>
            <TABLE cellSpacing=1 cellPadding=0 width=554 bgColor=#000000 
            border=0>
              <TBODY>
              <TR align=middle bgColor=#ffffff>
                <TD width=51>&nbsp;</TD>
                <TD width=27>b0</TD>
                <TD width=29>b1</TD>
                <TD width=31>b2</TD>
                <TD width=34>b3</TD>
                <TD width=132>FirstTarget</TD>
                <TD width=112>SecondTarget</TD>
                <TD width=129>ThirdTarget</TD></TR>
              <TR align=middle bgColor=#ffffff>
                <TD width=51>移动前</TD>
                <TD colSpan=4>D[0]</TD>
                <TD width=132>D[1]</TD>
                <TD width=112>D[2]</TD>
                <TD width=129>D[3]</TD></TR>
              <TR align=middle bgColor=#ffffff>
                <TD width=51>移动后</TD>
                <TD colSpan=4>D[1]</TD>
                <TD width=132>D[2]</TD>
                <TD width=112>D[3]</TD>
                <TD 
            width=129>D[0]</TD></TR></TBODY></TABLE><BR>b0和b2作为数组下标从S1中寻找s-box替换数:S1[b0],S1[b2]<BR>b1和b3作为数组下标从S0中寻找s-box替换数:S0[b1],S0[b3]<BR>对FirstTarget的操作:<BR>FirstTarget按位异或S1[b0]的结果返回给FirstTarget<BR>对SecondTarget的操作:<BR>SecondTarget加上S0[b3]的结果返回给SecondTarget<BR>对ThirdTarget的操作:<BR>ThirdTarget按位异或S0[b1]后加上S1[b2]的结果返回给ThirdTarget.<BR>本步骤共进行8轮,在第一轮和第五轮中操作结尾处添加将Source加上FirstTarget的结果返回给Source的操作.在第二轮和第六轮中操作结尾处添加将Source加上ThirdTarget的结果返回给Source的操作<BR><BR><B>2.3.2第二步密码核</B><BR>把输入的128位数据循环左移32位后分成四块D[0],D[1],D[2],D[3],其中D[0]作为源数据(Source),剩下的3个作为目标数据, 
            把Source循环右移13位的结果返回给Source,<BR>
            <TABLE cellSpacing=1 cellPadding=0 width=400 bgColor=#000000 

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