Need help with a C programming.
I have two matrixes Matrix A , 10X2 and Matrix B 10X2.<br/>I would like to achieve matrix C in the following format.
A=
1 1.24
2 1.49
0 0
0 0
3 5.69
4 3.78
5 2.8
6 9.2
0 0
0 0
B=
0 0
0 0
8 3.5
9 4.1
0 0
0 0
0 0
0 0
5 1.9
9 2.5
We are trying to operate these two matrices and achieve c, where c matrix looks like following, only mentioning the first colomn here
C=
9-8
9-8
0
0
9-5
9-5
9-5
9-5
0
0
Column 1 of C matrix is constructed like this. First, it will search the first column of B Matrix for a non-zero value, in the given example, it has found 8 and then 9 from the B Matrix. Then it will store 8 as low value, and 9 as high. It can be more values, but in this case, it was these two values. So after finding the low and high value it will put ‘High – Low’ in this case 9 – 8 and then it will search 1st column of Matrix A and replace with the ‘High – Low’ in this case 9 – 8, in the non zero places just before the group, as shown in Matrix C.
tried with the following code
Code:
#include <stdio.h><br/>
int main()<br/>
{<br/>
int B[]={0,0,8,9,0,0,0,0,5,9};<br/>
int A[]={1,2,0,0,3,4,5,6,0,0};<br/>
int n=0, x=0, y=0;
int high=0, low=0, result=0;
int i=0, j=0;
int C[n];
n=sizeof(B)/sizeof(B[0]);
printf("Length of B: %d \n",n);
//Print b
printf("Print B:");
for(i=0;i<n;i++)
{
printf("%d \t",B[i]);
}
printf("\n");
for(i=0;i<n;i++)
{
if(B[i]==0)
{
C[i]=B[i];
}
if((B[i]!=0)&&(i==0))
{
x=i;
low=B[0];
printf("Print low1:%d\n",low);
}
else if((B[i]==0)&&(B[i+1]!=0))
{
x=i+1;
low=B[i+1];
printf("Print low2:%d\n",low);
}<br/>
if((B[i]!=0)&&(B[i+1]==0))<br/>
{<br/>
y=i;<br/>
high=B[i];<br/>
printf("Print high1:%d\n",high);<br/>
result=high-low;
for(j=x;j<=y;j++)
{
C[j]=result;
}
}
else if((B[i]!=0)&&(i==n-1))
{
y=i;
high=B[i];
printf("Print high2:%d\n",high);
result=high-low;
for(j=x;j<y;j++)
{
C[j]=result;
}
}<br/>
if((B[i-1]==0)&&(B[i]!=0)&&(i==n-1))<br/>
{<br/>
C[i]=B[i];<br/>
}
}
//Print C
printf("Print C:");
for(i=0;i<n;i++)
{
printf("%d \t",C[i]);
}
printf("\n");
}
I am stuck a big time, any help will be appreciated