Answer:
He meant that you a USB drive in case your computer broke
Answer:
#include<stdio.h>
#include<conio.h>
int m=0,n=4;
int cal(int temp[10][10],int t[10][10])
{
int i,j,m=0;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(temp[i][j]!=t[i][j])
m++;
}
return m;
}
int check(int a[10][10],int t[10][10])
{
int i,j,f=1;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
if(a[i][j]!=t[i][j])
f=0;
return f;
}
void main()
{
int p,i,j,n=4,a[10][10],t[10][10],temp[10][10],r[10][10];
int m=0,x=0,y=0,d=1000,dmin=0,l=0;
clrscr();
printf("\nEnter the matrix to be solved,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&a[i][j]);
printf("\nEnter the target matrix,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&t[i][j]);
printf("\nEntered Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",a[i][j]);
printf("\n");
}
printf("\nTarget Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",t[i][j]);
printf("\n");
}
while(!(check(a,t)))
{
l++;
d=1000;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(a[i][j]==0)
{
x=i;
y=j;
}
}
//To move upwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=0)
{
p=temp[x][y];
temp[x][y]=temp[x-1][y];
temp[x-1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move downwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x+1][y];
temp[x+1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move right side
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x][y+1];
temp[x][y+1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move left
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=0)
{
p=temp[x][y];
temp[x][y]=temp[x][y-1];
temp[x][y-1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
printf("\nCalculated Intermediate Matrix Value :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",r[i][j]);
printf("\n");
}
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
a[i][j]=r[i][j];
temp[i][j]=0;
}
printf("Minimum cost : %d\n",d);
}
getch();
}
Explanation:
The correct answer is the Science, Technololy, Engineering and Math career cluster.
All of the five careers that are listed fall into the career cluster that is called Science, Technology, Engineering and Math. People working in this career cluster could work in science labs, design products and systems, or support scientists or mathematicians in their work.
