LOJ #570. 「LibreOJ Round #11」Misaka Network 与任务

观察发现，肯定是1个或两个最优。

#include<bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define db long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define IT iterator

#define PB push_back
#define MK make_pair
#define LB lower_bound
#define UB upper_bound
#define EB emplace_back
#define fi first
#define se second

#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define UPD(x,y) (((x)+=(y))>=mo?x-=mo:233)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));

#define LS3 k*2,l,mid
#define RS3 k*2+1,mid+1,r
#define LS5 k*2,l,mid,x,y
#define RS5 k*2+1,mid+1,r,x,y
#define GET pushdown(k);int mid=(l+r)/2

#define INF ((1ll<<60)-233)
#define sqr(x) ((x)*(x))
#define debug puts("wzpkking")
using namespace std;

const int mo=1000000007;
const int N=(1<<22)+5;
int f[N],cnt[N],n,m,k,ans;
int power(int x,int y){
    int s=1;
    for (;y;y/=2,x=1ll*x*x%mo)
        if (y&1) s=1ll*s*x%mo;
    return s;
}
int main(){
    scanf("%d%d%d",&n,&m,&k);
    for (int i=1,x;i<=m;i++)
        scanf("%d",&x),f[x]++;
    For(i,0,(1<<n)-1) cnt[i]=cnt[i>>1]+(i&1);
    For(i,0,n-1) For(j,0,(1<<n)-1)
        if (j&(1<<i)) f[j-(1<<i)]+=f[j];
    For(i,1,(1<<n)-1)
        if (cnt[i]&1) ans=(ans+power(f[i],k))%mo;
        else ans=(ans-power(f[i],k)+mo)%mo;
    printf("%d",ans);
}