Answer:
def power(base, expo):
if expo == 0:
return 1
else:
return base * power(base, expo-1)
Explanation:
*The code is in Python.
Create a method called power that takes base and expo as parameters
Check if the expo is equal to 0. If it is return 1 (This is our base case for the method, where it stops. This way our method will call itself "expo" times). If expo is not 0, return base * power(base, expo-1). (Call the method itself, decrease the expo by 1 in each call and multiply the base)
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.
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Hope I helped! :)
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