結果

問題 No.3171 Color Restoration
コンテスト
ユーザー 👑 binap
提出日時 2025-06-06 21:32:21
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 2,870 bytes
コンパイル時間 4,117 ms
コンパイル使用メモリ 262,088 KB
実行使用メモリ 7,844 KB
最終ジャッジ日時 2025-06-06 21:32:27
合計ジャッジ時間 5,198 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 26
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef pair<int, int> P;

template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}

template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}

int main(){
	vector<string> one = {"gray","brown","green","cyan","blue","yellow","orange","red"};
	vector<string> two = {"gray","green","blue","yellow","red"};
	vector<string> three = {"gray","green","cyan","blue","violet","orange","red"};
	
	vector<string> s(3);
	cin >> s;

	set<vector<string>> se;
	vector<int> perm = {0, 1, 2};
	do{
		int tmp = 0;
		if(count(one.begin(), one.end(), s[perm[0]]) == 1) tmp++;
		if(count(two.begin(), two.end(), s[perm[1]]) == 1) tmp++;
		if(count(three.begin(), three.end(), s[perm[2]]) == 1) tmp++;
		if(tmp == 3) se.insert(vector<string>{s[perm[0]], s[perm[1]], s[perm[2]]});
	}while(next_permutation(perm.begin(), perm.end()));
	
	if(int(se.size()) == 1) cout << "Yes\n";
	else cout << "No\n";
	
	return 0;
}
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