绝望了,INTEL编程大赛第一题。
为什么读取整个文件的开销要比读取整个文件且求出最大值的开销还要大呢?
我用的测试文件大小为1GB;内存1GB,CPU双核3.39G,硬盘STAT2接口160G
两个原代码如下:
// Float2.cpp
#include <stdio.h>
#include <time.h>
//#define _MY_DBG_
#ifdef _MY_DBG_
#define print(x) printf x;
#else
#define print(x) ;
#endif
#define DATA_PRE_COUNT (1024*4+3)
int main(int argc, char * argv[])
{
FILE * fp = NULL;
register float fMaxT;
register float fMin=0.0;
register float fMax=0.0;
float buf[DATA_PRE_COUNT];
int offset=0;
int baseOffset=0;
fp = fopen(argv[1], "rb");
clock_t c_start = clock();
int readlen = fread(buf, sizeof(float), DATA_PRE_COUNT, fp);
while (!feof(fp))
{
for (int i=0;i<DATA_PRE_COUNT-3;i++)
{
fMaxT = buf[i]*buf[i+1]*buf[i+2]*buf[i+3];
if (fMaxT>fMax)
{
fMax = fMaxT;
offset = baseOffset+i;
fMin=fMax/1e12f;
print(("[%9d] (%8.2f,%8.2f,%8.2f,%8.2f) <%8.2f>\n",
offset, buf[i],buf[i+1],buf[i+2],buf[i+3],fMax/1e12f));
}
}
baseOffset += DATA_PRE_COUNT-3;
buf[0] = buf[DATA_PRE_COUNT-3];
buf[1] = buf[DATA_PRE_COUNT-2];
buf[2] = buf[DATA_PRE_COUNT-1];
readlen = fread(buf+3, sizeof(float), DATA_PRE_COUNT-3, fp);
if (readlen != (DATA_PRE_COUNT-3))
{
int cnt = readlen-3;
if (cnt < 0)
{
break;
}
for (int j=0;j<cnt;j++)
{
fMaxT = buf[j]*buf[j+1]*buf[j+2]*buf[j+3];
if (fMaxT>fMax)
{
fMax = fMaxT;
offset = baseOffset+j;
fMin=fMax/1e12f;
print(("[%9d] (%8.2f,%8.2f,%8.2f,%8.2f) <%8.2f>\n",
offset, buf[j],buf[j+1],buf[j+2],buf[j+3],fMax/1e12f));
}
}
break;
}
}
clock_t c_end = clock();
printf("乘积=%f 首数的序号=%d\n", fMax, offset);
printf("clocks is %d\n", c_end-c_start);
return 0;
}
// Float3.cpp
#include <stdio.h>
#include <time.h>
//#define _MY_DBG_
#ifdef _MY_DBG_
#define print(x) printf x;
#else
#define print(x) ;
#endif
#define DATA_PRE_COUNT (1024*4+3)
int main(int argc, char * argv[])
{
FILE * fp = NULL;
register float fMaxT;
register float fMin=0.0;
register float fMax=0.0;
float buf[DATA_PRE_COUNT];
int offset=0;
int baseOffset=0;
fp = fopen(argv[1], "rb");
clock_t c_start = clock();
int readlen = fread(buf, sizeof(float), DATA_PRE_COUNT, fp);
while (!feof(fp))
{
/* for (int i=0;i<DATA_PRE_COUNT-3;i++)
{
fMaxT = buf[i]*buf[i+1]*buf[i+2]*buf[i+3];
if (fMaxT>fMax)
{
fMax = fMaxT;
offset = baseOffset+i;
fMin=fMax/1e12f;
print(("[%9d] (%8.2f,%8.2f,%8.2f,%8.2f) <%8.2f>\n",
offset, buf[i],buf[i+1],buf[i+2],buf[i+3],fMax/1e12f));
}
}
baseOffset += DATA_PRE_COUNT-3;
buf[0] = buf[DATA_PRE_COUNT-3];
buf[1] = buf[DATA_PRE_COUNT-2];
buf[2] = buf[DATA_PRE_COUNT-1];
*/
readlen = fread(buf+3, sizeof(float), DATA_PRE_COUNT-3, fp);
/* if (readlen != (DATA_PRE_COUNT-3))
{
int cnt = readlen-3;
if (cnt < 0)
{
break;
}
for (int j=0;j<cnt;j++)
{
fMaxT = buf[j]*buf[j+1]*buf[j+2]*buf[j+3];
if (fMaxT>fMax)
{
fMax = fMaxT;
offset = baseOffset+j;
fMin=fMax/1e12f;
print(("[%9d] (%8.2f,%8.2f,%8.2f,%8.2f) <%8.2f>\n",
offset, buf[j],buf[j+1],buf[j+2],buf[j+3],fMax/1e12f));
}
}
break;
}
*/
}
clock_t c_end = clock();
printf("乘积=%f 首数的序号=%d\n", fMax, offset);
printf("clocks is %d\n", c_end-c_start);
return 0;
}
Float2 data1.dat
乘积=9861133242466304.000000 首数的序号=110335127
clocks is 25812
Float3 data1.dat
乘积=0.000000 首数的序号=0
clocks is 26687
Float2 data1.dat
乘积=9861133242466304.000000 首数的序号=110335127
clocks is 26828
Float3 data1.dat
乘积=0.000000 首数的序号=0
clocks is 27140