Answer:
LAN
Explanation:
i am not sure but I think it is lan
Answer:
The correct option is A
Explanation:
In project management, earliest finish time for activity A refers to the earliest start time for succeeding activities such as B and C to start.
Assume that activities A and B comes before C, the earliest finish time for C can be arrived at by computing the earliest start-finish (critical path) of the activity with the largest EF.
That is, if two activities (A and B) come before activity C, one can estimate how long it's going to take to complete activity C if ones knows how long activity B will take (being the activity with the largest earliest finish time).
Cheers!
I think this is true but like think about the computer hardware and stuff
Explanation:
#include<iostream>
#include<string.h>
using namespace std;
char *removestring(char str[80])
{
int i,j,len;
len = strlen(str);
for( i = 0; i < len; i++)
{
if (str[i] == ' ')
{
for (j = i; j < len; j++)
str[j] = str[j+1];
len--;
}
}
return str;
}
int main ()
{
char str[80];
cout << "Enter a string : ";
cin.getline(str, 80);
strcpy(removestring(str), str);
cout << "Resultant string : " << str;
return 0;
}
In this program the input is obtained as an character array using getline(). Then it is passed to the user-defined function, then each character is analyzed and if space is found, then it is not copied.
C++ does not allow to return character array. So Character pointer is returned and the content is copied to the character array using "strcpy()". This is a built in function to copy pointer to array.
Answer:
The number of subproblems are given as 
while the value of T(n) to be less than S(n) is for 342.
Explanation:
The number of subproblems are given as
Asymptotic running time for Strassen’s algorithm is 
Now, when a increases, number of subproblems determines the asymptotic running time of the problem and case 1 of master theorem applies. So, in worst case, asymptotic running time of the algorithm will be
Now, for T(n) to be smaller than S(n)
So,
So,
a=342
