计数好题
容斥式子：发现只要每个钦定方案的贡献都考虑到再配上容斥系数就是对的
O(n^2)->O(n)
把麻烦的i=0，j=0特殊考虑下
剩下的，先把麻烦的东西化简干净
然后枚举一维i，剩下的二项式定理！！！！
#include<bits/stdc++.h> #define reg register int #define il inline #define fi first #define se second #define mk(a,b) make_pair(a,b) #define numb (ch^'0') #define pb push_back #define solid const auto & #define enter cout<<endl using namespace std; typedef long long ll; template<class T>il void rd(T &x){ char ch;x=0;bool fl=false; while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true); for(x=numb;isdigit(ch=getchar());x=x*10+numb); (fl==true)&&(x=-x); } template<class T>il void output(T x){if(x/10)output(x/10);putchar(x%10+'0');} template<class T>il void ot(T x){if(x<0) putchar('-'),x=-x;output(x);putchar(' ');} template<class T>il void prt(T a[],int st,int nd){for(reg i=st;i<=nd;++i) ot(a[i]);putchar('\n');} namespace Miracle{ const int N=1e6+5; const int mod=998244353; int ad(int x,int y){ return (x+y)>=mod?x+y-mod:x+y; } int qm(int x,ll y){ int ret=1; while(y){ if(y&1) ret=(ll)ret*x%mod;x=(ll)x*x%mod;y>>=1; }return ret; } int inv[N],jie[N]; int C(int n,int m){ if(n<0||m<0||n<m) return 0; return (ll)jie[n]*inv[m]%mod*inv[n-m]%mod; } int main(){ int n;rd(n); ll ans=0; jie[0]=1; for(reg i=1;i<=n;++i) jie[i]=(ll)jie[i-1]*i%mod; inv[n]=qm(jie[n],mod-2); for(reg i=n-1;i>=0;--i) inv[i]=(ll)inv[i+1]*(i+1)%mod; for(reg i=1;i<=n;++i){ ans=ad(ans,(i+1)&1?mod-(ll)C(n,i)*qm(3,(ll)n*(n-i)+i)%mod:(ll)C(n,i)*qm(3,(ll)n*(n-i)+i)%mod); } ans=ans*2%mod; ll sum=0; ll base=1; for(reg i=0;i<=n-1;++i){ sum=ad(sum,(i+1)&1?mod-(ll)C(n,i)*ad(qm(ad(1,mod-base),n),mod-qm(mod-base,n))%mod:(ll)C(n,i)*ad(qm(ad(1,mod-base),n),mod-qm(mod-base,n))%mod); base=base*3%mod; } ans=(ans+sum*3)%mod; ot(ans); return 0; } } signed main(){ Miracle::main(); return 0; } /* Author: *Miracle* */