jzoj5828 [省选模拟2018.8.18]⽔果拼盘 乱搞

Description

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好长啊

Solution

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正解是fwt一类的东西，好像还要生成函数。我的这种做法是乱搞得来的
考虑到期望的线性性，我们只需要考虑每个ai$a_i$的和bi$b_i$各自的贡献即可
我们记n个水果拼盘中第i个出现的次数为toti$to{t}_i$，那么ai$a_i$的系数显然为(nk)−(n−totik)(nk)$\frac{(\genfrac{}{}{0ex}{}{n}{k})-(\genfrac{}{}{0ex}{}{n-to{t}_i}{k})}{(\genfrac{}{}{0ex}{}{n}{k})}$，bi$b_i$的系数就是1-ai$a_i$的系数
然后预处理一波阶乘和逆元就行了

Code

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#include <stdio.h>
#include <string.h>
#include <bitset>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define drp(i,st,ed) for (int i=st;i>=ed;--i)

typedef long long LL;
const int MOD=998244353;
const int N=1100005;

int a[N],b[N],tot[N],n,m,k;

LL fac[N],inv[N],ans;

int read() {
    int x=0,v=1; char ch=getchar();
    for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):(v),ch=getchar());
    for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
    return x*v;
}

LL ksm(LL x,LL dep) {
    LL ret=1;
    for (;dep;) {
        if (dep&1) ret=ret*x%MOD;
        x=x*x%MOD; dep>>=1;
    }
    return ret;
}

LL C(int n,int m) {
    if (n<m) return 0;
    if (n==m) return 1;
    LL ret=fac[n]*inv[m]%MOD;
    ret=ret*inv[n-m]%MOD;
    return ret;
}

int main(void) {
    freopen("data.in","r",stdin);
    freopen("myp.out","w",stdout);
    n=read(),m=read(),k=read();
    fac[0]=1; rep(i,1,n) fac[i]=1LL*i*fac[i-1]%MOD;
    inv[n]=ksm(fac[n],MOD-2); drp(i,n-1,1) inv[i]=1LL*(i+1)*inv[i+1]%MOD;
    rep(i,1,m) a[i]=read();
    rep(i,1,m) b[i]=read();
    rep(i,1,n) {
        int s=read(),x;
        for (;s--;) {
            x=read();
            tot[x]++;
        }
    }
    rep(i,1,m) {
        LL tmp1=C(n,k);
        LL tmp2=C(n-tot[i],k);
        LL tmp3=(tmp1-tmp2+MOD)%MOD;
        ans=(ans+1LL*a[i]*tmp3%MOD*ksm(tmp1,MOD-2)%MOD)%MOD;
        ans=(ans+1LL*b[i]*tmp2%MOD*ksm(tmp1,MOD-2)%MOD)%MOD;
    }
    printf("%lld\n", ans);
    return 0;
}