板子

杂七杂八的数学题板子：

namespace Math{
	
	const int N=1e6+5;
	const int p=998244353;
	
	inline int mul(int a,int b){
		return ((1LL*((a%p+p)%p))*(1LL*((b%p+p)%p)))%p;
	}
	
	inline void mult(int &a,int b){
		return a=mul(a,b),void();
	}
	
	inline int qpow(int a,int b){
		int res=1;
		while(b){
			if(b&1)mult(res,a);
			mult(a,a),b>>=1;
		}
		return res;
	}
	
	inline int inv(int x){
		return qpow(x,p-2);
	}
	
	int mem[N]={1};
	
	inline int jc(int x){
		if(mem[x])return mem[x];
		return mem[x]=mul(x,jc(x-1));
	}
	
	inline int C(int n,int m){
		if(n<m)return 0;
		return mul(jc(n),inv(mul(jc(m),jc(n-m))));
	}
	
	int tot;
	
	int pri[N],phi[N],e[N];
	bool np[N];
	
	inline void init(int n){
		for(int i=2;i<=n;++i){
			if(!np[i])phi[i]=i-1,e[i]=pri[++tot]=i;
			for(int j=1;j<=tot&&i*pri[j]<=n;++j){
				np[i*pri[j]]=1,e[i*pri[j]]=pri[j];
				if(!(i%pri[j])){phi[i*pri[j]]=phi[i]*pri[j];break ;}
				phi[i*pri[j]]=phi[i]*(pri[j]-1);
			}
		}
		return phi[1]=1,void();
	}
	
	inline int qryphi(int x){
		int res=x;
		for(int i=2;i<=x/i;++i){
			if(x%i)continue ;
			(res*=(i-1))/=i;
			while(!(x%i))x/=i;
		}
		if(x>1)(res*=(x-1))/=x;
		return res;
	}
	
}using namespace Math;

FFT（迭代）：

namespace FFT{
	
	#define db double
	
	const int N=2e6+5;
	const db PI=acos(-1);
	
	int n,m;
	int lim=1,idx;
	
	int r[N<<1];
	
	struct cp{
		
		db x,y;
		
		friend cp operator + (const cp &a,const cp &b){
			return {a.x+b.x,a.y+b.y};
		}
		
		friend cp operator - (const cp &a,const cp &b){
			return {a.x-b.x,a.y-b.y};
		}
		
		friend cp operator * (const cp &a,const cp &b){
			return {a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};
		}
		
		cp& operator*=(const cp &b){
        	return *this=*this*b,*this;
    	}
		
	}A[N<<1],B[N<<1];
	
	inline void init(){
		while(lim<=n+m)lim<<=1,++idx;
		for(int i=0;i<lim;++i)r[i]=((r[i>>1]>>1)|((i&1)<<(idx-1)));
		return ;
	}
	
	inline void fft(cp *f,int op){
		for(int i=0;i<lim;++i)if(i<r[i])swap(f[i],f[r[i]]);
		for(int i=1;i<lim;i<<=1){
			cp wn={cos(PI/i),op*sin(PI/i)};
			for(int j=0;j<lim;j+=(i<<1)){
				cp w={1.0,0.0};
				for(int k=0;k<i;++k,w*=wn){
					cp y=f[j+k],z=w*f[i+j+k];
					f[j+k]=y+z,f[i+j+k]=y-z;
				}
			}
		}
		return ;
	}
	
	inline void solve(){
		n=read(),m=read();init();
		for(int i=0;i<=n;++i)A[i].x=read();
		for(int i=0;i<=m;++i)B[i].x=read();
		fft(A,1),fft(B,1);
		for(int i=0;i<lim;++i)A[i]*=B[i];
		fft(A,-1);
		for(int i=0;i<n+m+1;++i)printf("%d ",(int)(A[i].x/lim+0.5));
		return printf("\n"),void();
	}
	
}using namespace FFT;

NTT：

namespace NTT{
	
	const int N=2e6+5;
	const int p=998244353;
	const int g=3,inv_g=332748118;
	
	int n,m;
	int lim=1,idx;
	
	int A[N<<1],B[N<<1];
	int r[N<<1];
	
	inline int mul(int a,int b){
		return (1LL*a)*(1LL*b)%p;
	}
	
	inline void mult(int &a,int b){
		return a=mul(a,b),void();
	}
	
	inline int qpow(int a,int b){
		int res=1;
		while(b){
			if(b&1)mult(res,a);
			mult(a,a),b>>=1;
		}
		return res;
	}
	
	inline int inv(int x){
		return qpow(x,p-2);
	}
	
	inline void init(){
		while(lim<=n+m)lim<<=1,++idx;
		for(int i=0;i<lim;++i)r[i]=((r[i>>1]>>1)|((i&1)<<(idx-1)));
		return ;
	}
	
	inline void ntt(int *f,int op){
		for(int i=0;i<lim;++i)if(i<r[i])swap(f[i],f[r[i]]);
		for(int i=1;i<lim;i<<=1){
			int wn=qpow((~op)?g:inv_g,(p-1)/(i<<1));
			for(int j=0;j<lim;j+=(i<<1)){
				int w=1;
				for(int k=0;k<i;++k,mult(w,wn)){
					int y=f[j+k],z=mul(w,f[i+j+k]);
					f[j+k]=(y+z)%p,f[i+j+k]=(y-z+p)%p;
				}
			}
		}
		if(~op)return ;
		int inv_lim=inv(lim);
		for(int i=0;i<lim;++i)mult(f[i],inv_lim);
		return ;
	}
	
	inline void solve(){
		n=read(),m=read();init();
		for(int i=0;i<=n;++i)A[i]=read();
		for(int i=0;i<=m;++i)B[i]=read();
		ntt(A,1),ntt(B,1);
		for(int i=0;i<lim;++i)mult(A[i],B[i]);
		ntt(A,-1);
		for(int i=0;i<=n+m;++i)printf("%d ",A[i]);
		return printf("\n"),void();
	}
	
}using namespace NTT;