#include <stdio.h>
unsigned short a[8]={0x0180,0x0240,0x0660,0x0c30,0x1ff8,0x300c,0x6006,0xf00f};
int b[16];
void depress(unsigned short n)
{
int i;
unsigned short m;
for (i=15;i>=0;i--)
{
m=n>>i;
m&=0x1;
if (m)
b[15-i]=1;
else
b[15-i]=0;
}
}
int main()
{
int row,col,i;
printf("\n");
for (row=0;row<8;row++)
for (col=0;col<16;col++)
{
depress(a[row]);
if (b[col] ==0 )
printf(" ");
else
printf("*");
if (col==15)
printf("\n");
}
return 0;
}
还有就是怎么优化？

没必要单门depress，直接循环右移就行了

我看出来了，depress可以放到外层循环，这样总共只要执行8次
还有可以一次性把数据都depress完成，会更快一些
depress函数不知道怎么优化？

有人吗

可以去掉depress
A对称
内层循环中
if改成a［］& 1
a［］右移一位

话说把数组做大，应该可以显示汉字

重写了一边结果不对了
#include <stdio.h>
unsigned short a[8]={0x0180,0x0240,0x0660,0x0c30,0x1ff8,0x300c,0x6006,0xf00f};
struct UINT64
{
unsigned int low;
unsigned int high;
};
void main()
{
struct UINT64 begin,end;
int row,col;
_asm
{
rdtsc
mov begin.low,eax
mov begin.high,edx
}
/**************begin****************/
printf("\n");
for (row=0;row<8;row++)
{
for (col=0;col<16;col++)
{
if ( (a[row]>>(15-col))&0x1 == 0 )
printf(" ");
else
printf("*");
if (col==15)
printf("\n");
}
}
/*************end*********************/
_asm
{
rdtsc
mov end.low,eax
mov end.high,edx
}
printf("%010d,%010d ticks took\n",end.high-begin.high,end.low-begin.low);
}