int foo(int N) {
if (N == 0)
return 1;
else
return N * foo(N-1);
}
int bar(int N) {
if (N == 1)
return 1;
for (int i = 1; i < N; i++){
bar(N-1);
}
}
int bazo(int N) {
if (N < 1)
return 1;
for (int i = 1; i < N*N; i++){
for (j = 1; j < i; j++) {
bazo(N/2);
}
}
}
int hard(int N) {
if (N < 1)
return 1;
return (N*hard(N/2))/hard(N-1);
}
Any computer science student shoud be able to calculate hat:
foo(N) = T(N) = T(N-1) = T(N)
bar(N) = T(N) = (N-1)*T(N-1)
bazo(N) = T(N) = N2*O(N2)*T(N/2)
hard(N) = T(N) = T(N/2)+T(N-1)
No comments:
Post a Comment