A fuse protects computer system from voltage fluctuations
Answer:
#include <iostream>
using namespace std;
int main() {
int *ip_arr,n;//pointer name inp_arr and integer n to store the size.
cin>>n;//size.
for(int i=0;i<n;i++)
ip_arr[i]=-1;//assigning -1 to every element.
for(int i=0;i<n;i++)
{
cout<<ip_arr[i]<<" ";//printing every element.
}
return 0;
}
output:-
100
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
Explanation:
I am taking input of size.You should enter 100 for 100 values which have value -1.
Answer:
Number of strings = (10, 2)×(8,3) = 2520
Explanation:
The number of possible combinations for taking two 0's is C(10, 2)
It remains 8 Positions
The number of possible combinations for taking three 1's is C(8,3)
So there remains 5 spots
Answer:
public class Digits
{
public static boolean allDigitsOdd(int num)
{
boolean flag=true;
int rem;
while(num>0)
{
rem=num%10;
num=num/10;
if(rem%2==0) // if a even digit found immediately breaks out of loop
{
flag=false;
break;
}
}
return flag; //returns result
}
public static void main(String args[])
{
System.out.println(allDigitsOdd(1375)); //returns true as all are odd digits
}
}
OUTPUT :
true
Explanation:
Above program has 2 static methods inside a class Digits. Logic behind above function is that a number is divided by 10 until it is less than 0. Each time its remainder by 0 is checked if even immediately breaks out of the loop.
In terms of music taking one song and incorporating it into your own song
