[ARC145B] AB Game

The game is played by Alice and Bob. Initially, there are $n$ stones.

The players alternate turns, making a move described below, with Alice going first. The player who becomes unable to make a move loses.

• In Alice's turn, she must remove a number of stones that is a positive multiple of $A$.
• In Bob's turn, he must remove a number of stones that is a positive multiple of $B$.

In how many of Game $1$, Game $2$, ..., Game $N$ does Alice win when both players play optimally?

Constraints

• $1 \leq N ,A,B \leq 10^{18}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$ $A$ $B$

Output

Print the answer.

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Sample Input 1

4 2 1

Sample Output 1

2

In Game $1$, Alice cannot make a move and thus loses.

In Game $2$, Alice removes $2$ stones, and then Bob cannot make a move: Alice wins.

In Game $3$, Alice removes $2$ stones, Bob removes $1$ stone, and then Alice cannot make a move and loses.

In Game $4$, Alice removes $2 \times 2 = 4$ stones, and then Bob cannot make a move: Alice wins.

Therefore, Alice wins in two of the four games.

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Sample Input 2

27182818284 59045 23356

Sample Output 2

10752495144

### 题解 设这一局有 $x$ 个石子，先手取完石子后有 $x'$ 个石子。同时先手取得石子是 $a$ 的倍数，后手取的石子为 $b$ 的倍数 1. 若 $xa$且$a\le b$，先手可以取尽量多的石子，则$x'=x \bmod a$个石子，因为$x'=(x \bmod a)a$且$a>b$，如果先手取完石子后，$x'\ge b$，那么轮到后手时变成情况 2。所以当且仅当$x'综上所述，先手必胜的条件是 $(x\ge a)$ 且 $(x \bmod a<b)$

所以问题转化为有多少个 $x\in[A,n]$ 满足 $x \bmod A<b$

把 $[A,n]$ 的数按照 $\bmod A$ 的余数分组，每 $A$ 个为一组。那么共有 $\left\lfloor\frac{n-A}{A}\right\rfloor$ 个完整的组。每一组里面有 $\min(a,b)$ 个合法的数。剩下的 $n \bmod A +1$ 个数里面又有 $\min(n\bmod A+1,b)$ 个合法的数。我们就可以 $O(1)$ 求出答案。

注意特判 $n<A$

#include<bits/stdc++.h>
#define ll long long
#define p1 998244353
#define p2 1000000007
using namespace std;
ll n,a,b,ans;
int main()
{
	scanf("%lld%lld%lld",&n,&a,&b);
	if(n<a) printf("0");
	else{
		n-=a;
		ans+=n/a*min(a,b);
		n-=n/a*a;
		printf("%lld\n",ans+min(n+1,b));
	}
	return 0;
}