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Answer:
#include <iostream>
using namespace std;
int main() {
int a[4][5];//declaring a matrix of 4 rows and 5 columns.
for(int i=0;i<4;i++)
{
for(int j=0;j<5;j++)
{
if(i==3)//initializing last row as 0.
{
a[i][j]=0;
}
else//initializing last row as 1.
{
a[i][j]=1;
}
}
}
for(int i=0;i<4;i++)
{
for(int j=0;j<5;j++)
cout<<a[i][j]<<" ";//printing the matrix.
cout<<endl;
}
return 0;
}
Output:-
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
0 0 0 0 0
Explanation:
I have created a matrix of size 4 rows and 5 columns.I have used for loops to fill the array.To fill the last row with 0 i have used if statement.else we are filling it with 1.
<h2>This function will land up in infinite function call</h2>
Explanation:
first time when the function gets invoked,
f(6,8), so k=6 & n=8, inside the function it checks k==n, ie. 6==8, returns false, then one more if is available, so 6>8 is check for , once again it is false and else loop is executed, the function is called recursively using f(k-n,n), that is f(6-8,8), it means f(-2,8) is passed.
Second time,
if(-2==8) is false, so if(-2>8) is again false and function f(-10, 8) is called
if(-10==8) is false, so if(-10>8) is again false and function f(-18,8) is called
if(-18==8) is false, so if(-18>8) is again false and function f(-26,8) is called
So this goes recursively and ends in an infinite function call.
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