結果

問題 No.2322 MMA文字列
ユーザー ponjuiceponjuice
提出日時 2023-05-28 13:30:31
言語 C++23
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 5,328 bytes
コンパイル時間 5,466 ms
コンパイル使用メモリ 313,176 KB
実行使用メモリ 4,380 KB
最終ジャッジ日時 2023-08-27 07:47:20
合計ジャッジ時間 6,319 ms
ジャッジサーバーID
（参考情報） 		judge14 / judge11
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
4,376 KB
testcase_01 AC 1 ms
4,380 KB
testcase_02 AC 1 ms
4,376 KB
testcase_03 AC 2 ms
4,380 KB
testcase_04 AC 2 ms
4,376 KB
testcase_05 AC 2 ms
4,376 KB
testcase_06 AC 2 ms
4,376 KB
testcase_07 AC 2 ms
4,376 KB
testcase_08 AC 1 ms
4,376 KB
testcase_09 AC 1 ms
4,376 KB
testcase_10 AC 1 ms
4,380 KB
testcase_11 AC 2 ms
4,380 KB
testcase_12 AC 1 ms
4,376 KB
testcase_13 AC 2 ms
4,376 KB
testcase_14 AC 2 ms
4,376 KB
testcase_15 AC 1 ms
4,376 KB
testcase_16 AC 1 ms
4,376 KB
testcase_17 AC 2 ms
4,376 KB
testcase_18 AC 2 ms
4,380 KB
testcase_19 AC 1 ms
4,376 KB
testcase_20 AC 1 ms
4,376 KB
testcase_21 AC 1 ms
4,376 KB
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>
using namespace std;
#include <atcoder/all>
using namespace atcoder;

//高速化 
struct ponjuice{ponjuice(){cin.tie(0);ios::sync_with_stdio(0);cout<<fixed<<setprecision(20);}}PonJuice;
//#define endl '\n' //インタラクティブ問題の時は消す

//型
using ll = long long;
using ld = long double;
using mint = modint998244353;//1000000007;
template<class T>using vc = vector<T>; template<class T>using vvc = vc<vc<T>>; template<class T>using vvvc = vvc<vc<T>>;
using vi = vc<int>;  using vvi = vvc<int>;  using vvvi = vvvc<int>;
using vl = vc<ll>;   using vvl = vvc<ll>;   using vvvl = vvvc<ll>;
using pi = pair<int, int>;  using pl = pair<ll, ll>;
using ull = unsigned ll;
template<class T>using priq = priority_queue<T>;
template<class T>using priqg = priority_queue<T, vc<T>, greater<T>>;

// for文
#define overload4(a, b, c, d, e, ...) e
#define rep1(n)             for(ll i = 0; i < n; i++)
#define rep2(i, n)          for(ll i = 0; i < n; i++)
#define rep3(i, a, b)       for(ll i = a; i < b; i++)
#define rep4(i, a, b, step) for(ll i = a; i < b; i+= step)
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)
#define per1(n)             for(ll i = n-1; i >= 0; i--)
#define per2(i, n)          for(ll i = n-1; i >= 0; i--)
#define per3(i, a, b)       for(ll i = b-1; i >= a; i--)
#define per4(i, a, b, step) for(ll i = b-1; i >= a; i-= step)
#define per(...) overload4(__VA_ARGS__, per4, per3, per2, per1)(__VA_ARGS__)
#define fore1(a)       for(auto&& i : a)	
#define fore2(i,a)     for(auto&& i : a)
#define fore3(x,y,a)   for(auto&& [x, y] : a)
#define fore4(x,y,z,a) for(auto&& [x, y, z] : a)
#define fore(...) overload4(__VA_ARGS__, fore4, fore3, fore2, fore1)(__VA_ARGS__)

//関数
#define mp make_pair
#define mt make_tuple
#define a first
#define b second
#define pb emplace_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define si(x) (ll)(x).size()
template<class S, class T>inline bool chmax(S& a, T b){return a < b && ( a = b , true);}
template<class S, class T>inline bool chmin(S& a, T b){return a > b && ( a = b , true);}
template<class T>void uniq(vc<T>&a){sort(all(a));a.erase(unique(all(a)),a.end());}
template<class T>vc<T> operator++(vc<T>&v,signed){auto res = v;fore(e,v)e++;return res;}
template<class T>vc<T> operator--(vc<T>&v,signed){auto res = v;fore(e,v)e--;return res;}
template<class T>vc<T> operator++(vc<T>&v){fore(e,v)e++;return v;}
template<class T>vc<T> operator--(vc<T>&v){fore(e,v)e--;return v;}

//入出力(operator)
template<int T>istream&operator>>(istream&is,static_modint<T>&a){ll v;is>>v;a=v;return is;}
istream&operator>>(istream&is,modint&a){ll v;cin>>v;a=v;return is;}
template<class S,class T>istream&operator>>(istream&is,pair<S,T>&a){is>>a.a>>a.b;return is;}
template<class T>istream&operator>>(istream&is,vc<T>&a){fore(e,a)is>>e;return is;}

template<int T>ostream&operator<<(ostream&os,static_modint<T>a){return os<<a.val();}
ostream&operator<<(ostream&os,modint a){return os<<a.val();}
template<class S,class T>ostream&operator<<(ostream&os,pair<S,T>&a){return os<<a.a<<" "<<a.b;}
template<class T>ostream&operator<<(ostream&os,set<T>&a){fore(it,a){os<<it<<" ";}return os;}
template<class T>ostream&operator<<(ostream&os,multiset<T>&a){fore(it,a){os<<it<<" ";}return os;}
template<class S,class T>ostream&operator<<(ostream&os,map<S,T>&a){fore(x,y,a){os<<x<<" "<<y<<"\n";}return os;}
template<class T>ostream&operator<<(ostream&os,unordered_set<T>&a){fore(it,a){os<<it<<" ";}return os;}
template<class S,class T>ostream&operator<<(ostream&os,unordered_map<S,T>&a){fore(x,y,a){os<<x<<" "<<y<<"\n";}return os;}
template<class T>ostream&operator<<(ostream&os,vc<T>&a){fore(e,a)os<<e<<" ";return os;}
template<class T>ostream&operator<<(ostream&os,vvc<T>&a){fore(e,a)os<<e<<"\n";return os;}

//入出力(関数)
vi readvi(ll n){vi a(n);cin>>a;return a;}
vl readvl(ll n){vl a(n);cin>>a;return a;}
vvi readg(ll n,ll m,bool bidirected=true){vvi g(n);rep(i,m){ll a,b;cin>>a>>b;a--;b--;g[a].pb(b);if(bidirected)g[b].pb(a);}return g;}
vvc<pi>readgc(ll n,ll m,bool bidirected=true){vvc<pi> g(n);rep(i,m){ll a,b,c;cin>>a>>b>>c;a--;b--;g[a].pb(b,c);if(bidirected)g[b].pb(a,c);}return g;}
vvi readt(ll n,bool bidirected=true){return readg(n,n-1,bidirected);}
vvc<pi> readtc(ll n,bool bidirected=true){return readgc(n,n-1,bidirected);}

inline void yes(){cout << "Yes\n";}
inline void no(){cout << "No\n";}
inline void yesno(bool y = true){if(y)yes();else no();}
inline void print(){cout<<endl;}
template<class T>inline void print(T a){cout<<a<<endl;}
template<class T,class... Ts>inline void print(T a,Ts ...b){cout<<a;(cout<<...<<(cout<<' ',b));cout<<endl;}

//定数
constexpr ll mod = 998244353;
constexpr ll minf=-(1<<29);
constexpr ll inf=(1<<29);
constexpr ll MINF=-(1LL<<60);
constexpr ll INF=(1LL<<60);
constexpr ld EPS = 1e-8;
const ld PI = acosl(-1);
#define equals(a, b) (abs((a) - (b)) < EPS)
const int dx[4] ={-1, 0, 1, 0};
const int dy[4] ={ 0, 1, 0,-1};
const int dx8[8] ={-1,-1,-1, 0, 1, 1, 1, 0};
const int dy8[8] ={-1, 0, 1, 1, 1, 0,-1,-1};

void solve();
int main() {
	int t = 1;
    // cin >>t;
    while(t--)solve();
}

void solve(){
    int n,m;
    string s,t;
    cin >> s;
    if(s[0] == s[1] && s[0] != s[2]){
        yes();
    }else{
        no();
    }
}
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