Answer:
The answer is "Option a".
Explanation:
This design is the breakdown for a structure to smaller components to recognize the textural functionalities. This system analysis the built-in the top-down style, that defines and it does not describe some first-level components, which is often known as a staggered layout, and wrong choices can be described as follows:
- In option b, It is wrong because it is not used in web pages.
- In option c, It does not start with individual commands, that's why it is incorrect.
- In option d, It is wrong because it uses loops and classes, but it can't decompose the problem.
Answer:
There are two ways to print 1 to 1000
- Using Loops.
- Using Recursion.
Explanation:
Using loops
for(int i=1;i<=1000;i++)
{
cout<<i<<" ";
}
Using recursion
Remember you can implement recursion using a function only.
void print(int n)
{
if(n==0)
return;
print(n-1);
cout<<n<<" "';
}
you should pass 1000 as an argument to the function print.
Is it asking for a web site like .edu? or something like that?
Answer:
The program is written using PYTHON SCRIPT below;
N=int(input(" Enter number of Rows you want:"))
M=[] # this for storing the matrix
for i in range(N):
l=list(map(int,input("Enter the "+str(i+1)+" Row :").split()))
M.append(l)
print("The 2D Matrix is:\n")
for i in range(N):
print(end="\t")
print(M[i])
W=[] # to store the first non zero elemnt index
T=[] # to store that value is positive or negative
L=len(M[0])
for i in range(N):
for j in range(L):
if (M[i][j]==0):
continue
else:
W.append(j) # If the value is non zero append that postion to position list(W)
if(M[i][j]>0): #For checking it is positive or negative
T.append(+1)
else:
T.append(-1)
break
print()
print("The first Non Zero element List [W] : ",end="")
print(W)
print("Positive or Negative List [T] : ",end="")
print(T)
Explanation:
In order for the program to determine a set of test cases it takes in input of 2D matrix in an N numbet of rows.
It goes ahead to program and find the column index of the first non-zero value for each row in the matrix A, and also determines if that non-zero value is positive or negative. The If - Else conditions are met accordingly in running the program.
