abc333F - Bomb Game 2

abc333F - Bomb Game 2

设$f_{i,j}$表示在有i个人的队列中，第j个人成为第一个的概率。

$f_{n,1}=\frac{1}{2}{f}_{n,n}$
$f_{n,2}=\frac{1}{2}{f}_{n-1,1}+\frac{1}{2}{f}_{n,1}$
...
$f_{n,n}=\frac{1}{2}{f}_{n-1,n-1}+\frac{1}{2}{f}_{n-1,n-1}$

先解出$f_{n,n}$再回代即可。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
#include<queue>
#include<bitset>
#include<cmath>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (ll (i)=(b);(i)>=(a);(i)--)
#define mk(x,y) make_pair((x),(y))
#define A puts("Yes")
#define B puts("No")
using namespace std;
typedef double db;
typedef long long ll;
const int N=1e6+5;
const ll inf = 1ll << 60;
const ll mo = 998244353;
const ll mo1=1e9+7;
const ll mo2=1e9+9;
const ll P=131;
const ll Q=13331;
ll f[3005][3005],n,inv2,inv[N],b[N];
void add(ll &x,ll y){
	x=(x+y)%mo;
}
ll power(ll a,ll b){
	ll t=1,y=a%mo;
	while (b) {
		if (b&1) t=t*y%mo;
		y=y*y%mo;
		b/=2;
	}
	return t;
}
int main()
{
//	freopen("data.in","r",stdin);
	
	scanf("%lld",&n);
	
	inv2=-(mo/2);
	inv2+=mo;
	
	b[0]=1;
	fo(i,1,n) b[i]=b[i-1]*2ll%mo;
	
	inv[0]=1;
	fo(i,1,n) inv[i]=inv[i-1]*inv2%mo;
	
	f[1][1]=1;
	
	ll t=0;
	fo(i,2,n) {
		t=0;
		fo(j,1,i-1) {
			add(t, inv[j]*f[i-1][i-j]%mo);
		}
		
		f[i][i]=t*b[i]%mo*power(b[i]-1, mo-2)%mo;
		f[i][1]=inv2*f[i][i]%mo;
		
		fo(j,2,i-1) f[i][j]=(f[i-1][j-1]+f[i][j-1])%mo*inv2%mo;
	}
	
	fo(i,1,n) {
		f[n][i]=(f[n][i]%mo+mo)%mo;
		printf("%lld ",f[n][i]);
	}
	
	return 0;
}