如何优化重复除 [英] how to optimize repetitive addition
本文介绍了如何优化重复除的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
#define NULL 0
int main()
{
int *array1=NULL,*array2=NULL;
int x =add(array1[0],array2[0]);
int y =add(array1[1],array2[7]);
int x =add(array1[2],array2[3]);
int y =add(array1[3],array2[4]);
int x =add(array1[4],array2[6]);
int y =add(array1[5],array2[1]);
int x =add(array1[6],array2[5]);
int y =add(array1[7],array2[2]);
................
................
int x =add(array1[252],array2[0]);
int y =add(array1[253],array2[7]);
int x =add(array1[254],array2[3]);
int y =add(array1[255],array2[4]);
}
基本上指数数组1
由1递增开始从0到255为止
但数组2
该指数是从0到7所以我想这个优化多个另外固定。如何优化呢?
Basically index for array1
is incrementing by 1 starting from 0 to till 255
but the index for array2
is fixed from 0 to 7. So I want to optimize this multiple addition. How to optimize this?
推荐答案
你可以为
int j = 0,order[] = {0,7,3,4,6,1,5,2};
for(int i = 0;i <256; i +=2)
{
int x =add(array1[i],array2[order[j%8]]);
j++;
int y =add(array1[i+1],array2[order[j%8]]);
j++;
}
更新
另一种解决方案可以(如果你不想使用I + = 2)
UPDATE
alternate solution can be (if you want without using i+=2)
int j = 0,order[] = {0,7,3,4,6,1,5,2};
for(int i = 0;i <256; i ++)
{
int x =add(array1[i],array2[order[j%8]]);
j++;
i++;
if(i>=256) break; //Improves it if you have non even condition
int y =add(array1[i],array2[order[j%8]]);
j++;
}
编辑由Sam
现在我想基于x和y的值选择这两个值来比较比较
CurrentTre=256;
if (x > y)
{
*array3[0]= x;
*array4[CurrentTre +0] = 0;
}
else
{
*array3[i] = y;
*array4[CurrentTre + 0] = 1;
}
..........
..........
if (x > y)
{
*array3[0]= x;
*array4[CurrentTre +127] = 254;
}
else
{
*array3[i] = y;
*array4[CurrentTre + 127] = 255;
}
/////////////
my approach is this way
if (x > y)
{
*array3[i]= x;
*array4[int CurrentTre +i] = int number[i]<<1;
}
else
{
array3[i] = y;
array4[int CurrentTre + i] = int number[i]<<1|1;
}
} //end function main
我要优化code优化我下面给出
请检查是否我该做的正确与否..?
uint32 even_number[255] ={0};
uint32 loop_index1=0;
uint32 loop_index2=0;
uint16 order[256]={0,7,3,4,6,1,5,2,4,3,7,0,1,6,2,5,7,0,4,3,2,5,1,6,3,4,0
,7,6,1,5,2,4,3,7,0,1,6,2,5,0,7,3,4,5,2,6,1,3,4,0,7,6,1,5,2,7,0,4,3,2,5
,1,6,5,2,6,1,0,7,3,4,1,6,2,5,4,3,7,0,2,5,1,6,7,0,4,3,6,1,5,2,3,4,0,7,1
,6,2,5,4,3,7,0,5,2,6,1,0,7,3,4,6,1,5,2,3,4,0,7,2,5,1,6,7,0,4,3,3,4,0,7
,6,1,5,2,7,0,4,3,2,5,1,6,4,3,7,0,1,6,2,5,0,7,3,4,5,2,6,1,7,0,4,3,2,5,1
,6,3,4,0,7,6,1,5,2,0,7,3,4,5,2,6,1,4,3,7,0,1,6,2,5,6,1,5,2,3,4,0,7,2,5
,1,6,7,0,4,3,1,6,2,5,4,3,7,0,5,2,6,1,0,7,3,4,2,5,1,6,7,0,4,3,6,1,5,2,3
,4,0,7,5,2,6,1,0,7,3,4,1,6,2,5,4,3,7,0}; //all 256 values
for(loop_index1;loop_index1<256;loop_index1++)
{
m0= (CurrentState[loop_index1]+Branch[order[loop_index2]]);
loop_index2++;
loop_index1++;
if(loop_index1>=256)
break;
m1= (CurrentState[loop_index1]+Branch[order[loop_index2]]);
loop_index2++;
if (mo > m1)
{
NextState[loop_index1]= m0;
SurvivorState[CurrentTrellis + loop_index1] =
even_number[loop_index1]<<1;
}
else
{
NextState[loop_index1] = StateMetric1;
SurvivorState[CurrentTrellis + loop_index1] =
even_number[loop_index1<<1|1;
}
}
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