int bob(longlongint r[], int t, int n)
{
int s = 0, c, i;
for (i = n - 1; i >= 0; i--) {
if ((r[i] % t) == 0) {
s = i + 1;
break;
}
}
if (s == 0)
return 0;
else {
for (c = s - 1; c < n - 1; c++) {
r[c] = r[c + 1];
}
}
return (n - 1);
}
int main()
{
int T;
cin >> T;
int a, b, n, i;
while (T--) {
cin >> n >> a >> b;
longlongint r[n];
for (i = 0; i < n; i++) {
cin >> r[i];
}
while (1) {
if ((n = bob(r, a, n)) == 0) {
cout << "ALICE\n";
break;
}
if ((n = bob(r, b, n)) == 0) {
cout << "BOB\n";
break;
}
}
}
return 0;
}
> In each turn, the current player must remove a non-zero number of elements from the sequence;
Have you considered whether removing more than one at once always gives the same answer?
What about removing all the multiples on the first attempt?
> long long int r[n];
This isn't valid C++.
1 2 3 4 5 6 7 8 9 10 11
$ g++ -pedantic -Wall -Wextra main.cpp
main.cpp:3:14: warning: ISO C++ 1998 does not support ‘longlong’ [-Wlong-long]
int bob(longlongint r[], int t, int n)
^
main.cpp: In function ‘int main()’:
main.cpp:29:10: warning: ISO C++ 1998 does not support ‘longlong’ [-Wlong-long]
longlongint r[n];
^
main.cpp:29:22: warning: ISO C++ forbids variable length array ‘r’ [-Wvla]
longlongint r[n];
^
You don't need long long, your input values only go up to 10^{9}, which fits in a long anyway.
> r[c] = r[c + 1];
You don't need to shuffle the whole array down one spot.
Just move the last element to the slot to be deleted. r[s-1] = r[n];
Basically, it's a race to see who can claim the 6 (being a common multiple of both 2 and 3).
There are two phrases in the challenge which seem interesting.
> both Bob and Alice play optimally.
> the current player must remove a non-zero number of elements from the sequence;
It's not a question of removing the first multiple of 2 or 3 from the sequence, but removing the "best" number from the sequence (in this case 6).
Alice can win if Bob is stupid and chooses 3 first, and she plays smart and picks 6 (leaving 2 4 9).
There is no point in Bob picking 3 or 9 early in the game, because these numbers can never be chosen by Alice. Similarly for Alice, 2 and 4 can be left until late in the game.
can't we just take all multiples of bob lucky number and remove them ??
in your example provided if there was also 12 can bob remove 3,9,6,12 in single turn
can't we just take all multiples of bob lucky number and remove them ??
in your example provided if there was also 12 can bob remove 3,9,6,12 in single turn
Yes you could, but then Bob would lose the next turn because he would be without moves.