题解 按位或

传送门

考场上完全没思路，瞎剪枝还把自己剪死了

• 整除也是可以按位拆开考虑的，第 \(i\) 位在剩余系下的贡献为 \(x\equiv 2^i \pmod p\)

首先如果没有那个能被3整除的限制就把 \(t\) 中为1的位抽出来容斥做就可以了
但现在有个每个数都要能被3整除的限制
先看原来的式子

\[\sum\limits_{i=1}^{cnt}(-1)^i\binom{cnt}{i}(2^{cnt-i})^n \]

考虑怎么在容斥的时候加上整除的限制
发现一个事情：

\[2^i \equiv \begin{cases} 2\pmod 3&{2\nmid i} \\1 \pmod 3&{2\mid i}\end{cases} \]

于是要求容斥的时候为1的位的总贡献在模3的意义下等于0
先把贡献不同的位分开，令 \(odd\) 为奇位的个数，\(even\) 为偶位的个数
于是拆一下式子，\(\binom{cnt}{i} = \sum\limits_{k=0}^i\binom{odd}{k}\binom{even}{i-k}\)
考虑后面那部分怎么算
可以直接背包算出总贡献在模3的意义下等于0的选法数
于是复杂度 \(O(log^3n)\)

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
#define pb push_back
// #define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline ll read() {
	ll ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

ll n, t;
const ll mod=998244353;
inline void md(ll& a, ll b) {a+=b; a=a>=mod?a-mod:a;}

namespace force{
	ll dp[110][150];
	void solve() {
		dp[0][0]=1;
		for (int i=1; i<=n; ++i) {
			for (int j=t; ~j; --j) if (j%3==0 && (j&t)==j) {
				for (int k=0; k<=t; ++k) {
					md(dp[i][j|k], dp[i-1][k]);
				}
			}
		}
		printf("%lld\n", dp[n][t]);
		exit(0);
	}
}

namespace force2{
	ll dp[3100][3100];
	void solve() {
		dp[0][0]=1;
		for (int i=1; i<=n; ++i) {
			for (int j=t; ~j; --j) if (j%3==0 && (j&t)==j) {
				for (int k=0; k<=t; ++k) {
					md(dp[i][j|k], dp[i-1][k]);
				}
			}
		}
		printf("%lld\n", dp[n][t]);
		exit(0);
	}
}

namespace task1{
	struct matrix{
		int n, m;
		int a[130][130];
		matrix(){memset(a, 0, sizeof(a));}
		matrix(int x, int y) {n=x; m=y; memset(a, 0, sizeof(a));}
		inline void resize(int x, int y) {n=x; m=y; memset(a, 0, sizeof(a));}
		inline void put() {for (int i=1; i<=n; ++i) {for (int j=1; j<=m; ++j) cout<<a[i][j]<<' '; cout<<endl;}}
		inline int* operator [] (int t) {return a[t];}
		inline matrix operator * (matrix b) {
			matrix ans(n, b.m);
			for (int i=1; i<=n; ++i)
				for (int k=1; k<=m; ++k)
					for (int j=1; j<=b.m; ++j)
						ans[i][j]=(ans[i][j]+1ll*a[i][k]*b[k][j])%mod;
			return ans;
		}
	}mat, tran;
	inline matrix qpow(matrix a, ll b) {matrix ans=a; --b; for (; b; a=a*a,b>>=1) if (b&1) ans=ans*a; return ans;}
	void solve() {
		mat.resize(1, t+1); tran.resize(t+1, t+1);
		mat[1][0+1]=1;
		for (int j=t; ~j; --j) if (j%3==0 && (j&t)==j) {
			for (int k=0; k<=t; ++k) {
				tran[k+1][(j|k)+1]+=1;
			}
		}
		mat=mat*qpow(tran, n);
		printf("%d\n", mat[1][t+1]);
		exit(0);
	}
}

namespace task{
	int cnt, odd, even;
	ll fac[N], inv[N], ans, f[3], g[3];
	inline ll C(int n, int k) {return n<k?0:fac[n]*inv[k]%mod*inv[n-k]%mod;}
	inline ll qpow(ll a, ll b) {ll ans=1; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}
	void solve() {
		fac[0]=fac[1]=1; inv[0]=inv[1]=1;
		for (int i=2; i<70; ++i) fac[i]=fac[i-1]*i%mod;
		for (int i=2; i<70; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
		for (int i=2; i<70; ++i) inv[i]=inv[i-1]*inv[i]%mod;
		for (int i=0; i<64; ++i) if (t&(1ll<<i)) ++cnt, ++(i&1?odd:even);
		for (int i=0; i<=cnt; ++i) {
			ll tem=0;
			for (int k=0; k<=i; ++k) {
				memset(f, 0, sizeof(f));
				f[0]=1;
				for (int s=1; s<=odd-k; ++s) {
					for (int t=2; ~t; --t) g[t]=f[t];
					for (int t=2; ~t; --t) g[(t+2)%3]=(g[(t+2)%3]+f[t])%mod;
					for (int t=2; ~t; --t) f[t]=g[t];
				}
				for (int s=1; s<=even-i+k; ++s) {
					for (int t=2; ~t; --t) g[t]=f[t];
					for (int t=2; ~t; --t) g[(t+1)%3]=(g[(t+1)%3]+f[t])%mod;
					for (int t=2; ~t; --t) f[t]=g[t];
				}
				// cout<<"f0: "<<f[0]<<endl;
				tem=(tem+C(odd, k)*C(even, i-k)%mod*qpow(f[0], n))%mod;
			}
			ans=(ans+(i&1?-1:1)*tem)%mod;
		}
		printf("%lld\n", (ans%mod+mod)%mod);
		exit(0);
	}
}

signed main()
{
	freopen("or.in", "r", stdin);
	freopen("or.out", "w", stdout);

	n=read(); t=read();
	// if (t<=100) task1::solve();
	// else force2::solve();
	task::solve();

	return 0;
}