CF1365G Secure Password

Solution

妙妙思维题。

注意到$\left(\genfrac{}{}{0ex}{}{13}{6}\right)>n$。（谁tm能注意到这个？！？）

我们可以把所有13位二进制数中有6个1的拿出来给$n$个数重标号。然后对于每一位$i$，求出重标号后这一位是1的数的位或和，记为$w_i$。

对于重标号后为$i$的位置的答案，只需要让所有$i$的0的位的$w_j$或起来即可。

Code

#include <bits/stdc++.h>using namespace std;template<typename T> inline bool upmin(T &x, T y) { return y < x ? x = y, 1 : 0; }
template<typename T> inline bool upmax(T &x, T y) { return x < y ? x = y, 1 : 0; }#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondtypedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int, int> PR;
typedef vector<int> VI; const lod eps = 1e-9;
const lod pi = acos(-1);
const int oo = 1 << 30;
const ll loo = 1ll << 60;
const int mods = 1e9 + 7;
const int inv2 = (mods + 1) >> 1;
const int MAXN = 1005;
const int INF = 0x3f3f3f3f; //1061109567
/*--------------------------------------------------------------------*/ll a[5005], w[15], num = 0;
vector<ll> V;
signed main() {
#ifndef ONLINE_JUDGE
//	freopen("a.in", "r", stdin);
#endifint n;cin >> n;for (int i = 0; i < 1 << 13 && num < n ; ++ i)if (__builtin_popcount(i) == 6) a[++ num] = i;for (int i = 0; i < 13 ; ++ i) {vector<int> V; V.clear();for (int j = 1; j <= n ; ++ j)if ((a[j] >> i) & 1) V.PB(j);if (!V.size()) continue;cout << "? " << V.size();for (auto v : V) cout << " " << v; cout << endl;cin >> w[i];}cout << "!";for (int i = 1; i <= n ; ++ i) {ll ans = 0;for (int j = 0; j < 13 ; ++ j)if (!((a[i] >> j) & 1)) ans |= w[j];cout << " " << ans;}return 0;
}