CF997C Sky Full of Stars

CF997C Sky Full of Stars 

计数好题

在Ta的博客查看

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('
');}

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*
*/