結果
問題 | No.2090 否定論理積と充足可能性 |
ユーザー |
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提出日時 | 2023-04-08 22:49:47 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 1,984 bytes |
コンパイル時間 | 5,671 ms |
コンパイル使用メモリ | 317,784 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-10-03 20:07:56 |
合計ジャッジ時間 | 6,663 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
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ファイルパターン | 結果 |
---|---|
other | AC * 20 |
ソースコード
#include <bits/stdc++.h>using namespace std;#include <atcoder/all>using namespace atcoder;using ll = long long;using ld = long double;using P = pair<int, int>;using Graph = vector<vector<ll>>;using vi = vector<int>;using vl = vector<long>;using vll = vector<long long>;using vvi = vector<vi>;using vvl = vector<vl>;using vvll = vector<vll>;using vs = vector<string>;using vc = vector<char>;using vvc = vector<vc>;using pll = pair<long long, long long>;using vpll = vector<pll>;using mint = modint1000000007;const long double EPS = 1e-18;const long long INF = 1e18;const long double PI = acos(-1.0L);#define reps(i, a, n) for (ll i = (a); i < (ll)(n); i++)#define rep(i, n) for (ll i = (0); i < (ll)(n); i++)#define rrep(i, n) for (ll i = (1); i < (ll)(n + 1); i++)#define repd(i, n) for (ll i = n - 1; i >= 0; i--)#define rrepd(i, n) for (ll i = n; i >= 1; i--)#define ALL(n) begin(n), end(n)#define IN(a, x, b) (a <= x && x < b)#define INIT \std::ios::sync_with_stdio(false); \std::cin.tie(0);template <class T>inline T CHMAX(T& a, const T b) {return a = (a < b) ? b : a;}template <class T>inline T CHMIN(T& a, const T b) {return a = (a > b) ? b : a;}ll NAND(ll a, ll b) { return ~(a & b); }ll gate(vll A) {return NAND((NAND((NAND(A[0], A[1])), A[2])),(NAND((NAND(A[3], A[4])), A[5])));}int main() {vll A;map<string, ll> ma;ll res = 0;rep(i, 6) {string s;cin >> s;if (!ma.count(s)) {ma[s] = res;res++;}A.push_back(ma[s]);}ll ans = 0;rep(bit, (1ll << 6)) {vll B;rep(i, 6) {if (bit & (1ll << A[i])) {B.push_back(1);} else {B.push_back(0);}}CHMAX(ans, gate(B));}if (ans) {cout << "YES" << endl;} else {cout << "NO" << endl;}}