Answer:
The solution is as follows.
class LFilters implements Lock {
int[] lvl;
int[] vic;
public LFilters(int n, int l) {
lvl = new int[max(n-l+1,0)];
vic = new int[max(n-l+1,0)];
for (int i = 0; i < n-l+1; i++) {
lvl[i] = 0;
}
}
public void lock() {
int me = ThreadID.get();
for (int i = 1; i < n-l+1; i++) { // attempt level i
lvl[me] = i;
vic[i] = me;
// rotate while conflicts exist
int above = l+1;
while (above > l && vic[i] == me) {
above = 0;
for (int k = 0; k < n; k++) {
if (lvl[k] >= i) above++;
}
}
}
}
public void unlock() {
int me = ThreadID.get();
lvl[me] = 0;
}
}
Explanation:
The code is presented above in which the a class is formed which has two variables, lvl and vic. It performs the operation of lock as indicated above.
Answer:
There is no need to make an algorithm for this simple problem. Just add the two numbers by storing in two different variables as follows:
Let a,b be two numbers.
c=a+b;
print(c);
But, if you want to find the sum of more numbers, you can use any loop like for, while or do-while as follows:
Let a be the variable where the input numbers are stored.
while(f==1)
{
printf(“Enter number”);
scanf(“Take number into the variable a”);
sum=sum+a;
printf(“Do you want to enter more numbers? 1 for yes, 0 for no”);
scanf(“Take the input into the variable f”);
}
print(Sum)
Explanation:
hi there answer is given mar me as brainliest
Float: 26.2
Array: c, o, m, p, u, t, er
Boolean: false
Character: c
