动物森友会

第一行包含 2 个整数 n, e ( $1 \leq n \leq 1\,000$ , $1 \leq e \leq 100$ )，中间以空格分隔，分别表示不同事件的数量和 Compute 一天能完成的事件次数。
接下来 n 行，每行包含 $m_i + 2$ 个整数 $c_i$ , $m_i$ , $a_{i_1}, a_{i_2}, \ldots, a_{i_{m_i}}$ ( $1 \le c_i \le 10 ^ 5$ , $1 \leq m_i \leq 7$ , $1 \leq a_{i_j} \leq 7$ , $\forall j\ne k, a_{i_j}\ne a_{i_k}$ )，中间以空格分隔，分别表示事件 i 需要完成的次数、事件开放的天数和事件 i 在一周中的哪些时间会开放。

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import java.util.Scanner;
public class Main {
    public static void main(String[] arg) {
        Scanner scanner = new Scanner(System.in);
        // todo
    }
}
 
 
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#include<bits/stdc++.h>

using namespace std;

using ll = long long;

using pii = pair<int, int>;

using db = double;

using vi = vector<int>;

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=n-1;i>=a;i--)

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define SZ(x) ((int)(x).size())

#define debug(x) cout << #x <<"\t" << x <<endl

const int inf = 0x3f3f3f3f;

const db eps = 1e-8;

const int mod = 1e9+7;

ll qpow(ll a, ll b){

ll ret = 1;

while(b){

if(b&1)ret = ret*a%mod;

a = a*a%mod;

b>>=1;

}

return ret;

}

int n, e;

int sta[1010];

int c[1010], num[10];

bool check(int x){

memset(num, 0, sizeof(num));

for(int i=1; i<=7; ++i) num[i] = x/7;

for(int i=1; i<=x%7; ++i) num[i]++;

for(int msk = 0; msk<(1<<7); msk++){

ll R=0;

for(int i=1; i<=7; ++i) if(msk&(1<<(i-1)))R += 1ll*num[i]*e;

ll L = 0;

for(int i=1; i<=n; ++i) if((sta[i]&msk)==sta[i]) L += c[i];

if(L>R)return 0;

}

return 1;

}

int main()

{

ios::sync_with_stdio(false);

cin.tie(0);

cin >> n >> e;

for(int i=1; i<=n; ++i){

cin >> c[i];

int m; cin >> m;

int msk = 0;

while(m--){

int x;cin>> x;

msk |= (1<<(x-1));

}

sta[i] = msk;

}

int l = 1, r = 700000000;

while(l<=r){

int m = l+r>>1;

if(check(m)) r = m-1;

else l = m+1;

// debug(m);

}

cout << l <<"\n";

}

#include<bits/stdc++.h>

using namespace std;

int main()

{

int i,j,k,x,n,e,ans=0,R[135]={0};

scanf("%d%d",&n,&e);

while(n--)

{

bool V[7]={0};

scanf("%d%d",&x,&k);

for(i=0;i<k;i++)scanf("%d",&j),V[j-1]=1;

for(i=1;i<128;i++)

{

for(j=0;j<7;j++)if(V[j]&&!((1<<j)&i))break;

if(j==7)R[i]+=x;

}

for(i=1;i<(1<<7);i++)

{

if(!R[i])continue;

n=__builtin_popcount(i),k=R[i]/e+(R[i]%e?1:0);

for(j=0;j<7;j++)

{

if(!((1<<j)&i))continue;

if((--k)%n==0)break;

}

ans=max(ans,k/n*7+j+1);

}

printf("%d\n",ans);

return 0;

}

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