牛客挑战赛43 D-数组操作

题目描述

题解

因为晚上摸鱼去了所以没打

设i有p个0q个1，则i的EGF（乘上(p+q)后）为

(A=0:pe^{p+q}-p)

(A=1:pe^{p+q}+q)

分治卷起来之后求x^k即可，特判p=q=0

code

#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define mod 998244353
#define Mod 998244351
#define G 3
#define ll long long
//#define file
using namespace std;

int c[100001],d[100001][2],N,N2,len,n,m,n1,n2,K,i,j,k,l;
ll A[131072],a[131072],b[131072],ans,s;
vector<ll> f[100001];

struct type{int s,id;} s1,s2;
bool operator < (type a,type b) {return a.s>b.s;}
priority_queue<type> hp;

ll qpower(ll a,int b) {ll ans=1; while (b) {if (b&1) ans=ans*a%mod;a=a*a%mod;b>>=1;} return ans;}
void dft(ll *a,int tp)
{
	int i,j,k,l,S=N,s1=2,s2=1;
	ll u,v,w,W;
	
	fo(i,0,N-1)
	{
		j=i,k=0;
		fo(l,1,len)
		k=k*2+(j&1),j>>=1;
		A[i]=a[k];
	}
	memcpy(a,A,N*8);
	
	fo(i,1,len)
	{
		w=(tp==1)?qpower(G,(mod-1)/s1):qpower(G,(mod-1)-(mod-1)/s1),S>>=1;
		fo(j,0,S-1)
		{
			W=1;
			fo(k,0,s2-1)
			{
				u=a[j*s1+k],v=a[j*s1+k+s2]*W;
				a[j*s1+k]=(u+v)%mod;
				a[j*s1+k+s2]=(u-v)%mod;
				W=W*w%mod;
			}
		}
		s1<<=1,s2<<=1;
	}
}

int main()
{
	#ifdef file
	freopen("d.in","r",stdin);
	#endif
	
	scanf("%d%d%d",&n,&m,&K),len=ceil(log2(n)),N=qpower(2,len);
	fo(i,1,n) scanf("%d",&c[i]);
	fo(i,1,m) scanf("%d%d",&j,&k),++d[j][k];
	
	fo(i,1,n)
	{
		if (d[i][0])
		{
			fo(j,0,d[i][0]+d[i][1]) f[i].push_back(0);
			f[i][d[i][0]+d[i][1]]=d[i][0];
			f[i][0]=(c[i])?(-d[i][0]):d[i][1];
			l=d[i][0]+d[i][1];
		}
		else
		if (c[i]) {printf("0
");exit(0);}
		else
		f[i].push_back(d[i][1]),l=0;
		
		if (d[i][0] || d[i][1])
		hp.push({l,i});
	}
	
	while (hp.size()>1)
	{
		s1=hp.top(),hp.pop();
		s2=hp.top(),hp.pop();
		
		n1=s1.s,n2=s2.s;
		len=ceil(log2(n1+n2+1)),N=qpower(2,len),N2=qpower(N,Mod);
		memset(a,0,N*8),memset(b,0,N*8);
		fo(i,0,n1) a[i]=f[s1.id][i];
		fo(i,0,n2) b[i]=f[s2.id][i];
		
		dft(a,1),dft(b,1);
		fo(i,0,N-1) a[i]=a[i]*b[i]%mod;
		dft(a,-1);
		fo(i,0,N-1) a[i]=a[i]*N2%mod;
		
		s1.s=n1+n2;
		fo(i,0,s2.s) f[s1.id].push_back(0);
		fo(i,0,s1.s) f[s1.id][i]=a[i];
		hp.push(s1);
	}
	s1=hp.top();
	fo(i,0,s1.s) s=f[s1.id][i],s=(s+mod)%mod,ans=(ans+s*qpower(i,K))%mod;
	
	fo(i,1,n) if (d[i][0]+d[i][1]) ans=ans*qpower(d[i][0]+d[i][1],Mod)%mod;
	printf("%lld
",(ans+mod)%mod);
	
	fclose(stdin);
	fclose(stdout);
	return 0;
}