逻辑优化方法

#define STAGE 6
#define Num (1<<STAGE*2)
#define Num4 (Num>>2)
#define Num16 (Num4>>2)
const int e = 1 << (2*k);
for(j=0;j<Num16;j++)
{
{
for(i=0;i<4;i++)
{
a=((4*(i+4*j))%Num*e)%Num/e;
}
}
}

[此贴子已经被作者于2007-11-12 13:33:43编辑过]

----------------解决方案--------------------------------------------------------
#define STAGE 6
#define Num (1<<STAGE*2)

Num (1 << 6 * 2) // Num (1 << 12) 即 1 * 2的12次方
----------------解决方案--------------------------------------------------------
我本来就是向得到 Num = 4 ^ STAGE 呀

----------------解决方案--------------------------------------------------------
a=((4*(i+4*j))*e)%n/e;发现可以去掉个 %n
----------------解决方案--------------------------------------------------------
继续改进如下
#define STAGE 2
#define Num (1<<STAGE*2)
#define Num4 (Num>>2)
#define Num16 (Num4>>2)
const int e = Num >> (2*k);

for(j=0;j<Num16;j++)
{
for(i=0;i<4;i++)
{
a= (4*(i+4*j))%e;
}
}
----------------解决方案--------------------------------------------------------