#include<stdio.h>
#include<stdlib.h>
#include <cstdlib>
#include<math.h>
using namespace std;
//定义文件路径
#define F_Path1 "F:\featuredat+.dat"
#define F_Path2 "F:\featuredat+de.dat"
#define N1 44564
#define N2 963
int main()
{
int *feature1;
feature1=(int*)calloc(N1*128,sizeof(int));
int *feature2;
feature2=(int*)calloc(N2*128,sizeof(int));
int i,j,k,pv,px,t[N2];
float min;
double temp;
//定义文件指针
FILE * fp1;
FILE * fp2;
//假如文件open失败
if((fp1=fopen(F_Path1,"rb"))==NULL)
{ printf("\nCan"t open file");
exit(0);
}
if ((fp2=fopen(F_Path2,"rb"))==NULL)
{
printf("\nCan"t open file");
exit(0);
}
fread(feature1,sizeof(int),128*N1,fp1);
fread(feature2,sizeof(int),128*N2,fp2);
//文件打开成功
px=0L;pv=0L;
min=0.0;
for(i=0;i<N2;i++)
{
for(j=0;j<N1;j++)
{
temp=0.0;
for(k=0;k<128;k++)
temp+=(float)pow((float)(feature1[px+k]-feature2[pv+k]),(int)2);
temp/=128.000000;
if((i==0)||(temp<min))
{
min=temp;
t[i]=int(j);
}
px+=128;
}
printf("%d ",t[i]);
pv+=128;
px=0L;
}
free(feature1);
free(feature2);
}
解决方案
50
也许你的数据就该得出这样的结果?
50
推荐使用WinHex软件查看硬盘或文件或内存中的原始字节内容。