結果

問題 No.2063 ±2^k operations (easy)
ユーザー lbm364dl
提出日時 2022-09-16 17:01:47
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 2,000 ms
コード長 2,656 bytes
コンパイル時間 775 ms
コンパイル使用メモリ 76,072 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-12-21 13:53:55
合計ジャッジ時間 1,864 ms
ジャッジサーバーID
（参考情報） 		judge5 / 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 * 21
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 #
プレゼンテーションモードにする

//#include <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
//#include <ext/pb_ds/assoc_container.hpp>
//using namespace __gnu_pbds;
using namespace std;
#define show(x) cerr << #x" = " << (x) << "\n"
#define pb push_back
#define pp pop_back
#define mp make_pair
#define fst first
#define snd second
#define FOR(var, from, to) for(int var = from; var < int(to); ++var)
#define all(x) x.begin(), x.end()
#define rev(x) x.rbegin(), x.rend()
#define sz(x) int(x.size())
#define vec(x) vector<x>
#define INF 2000000000
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
// use unique second element of pair to work as multiset
//typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
const ll mod = 1e9 + 7;
template<typename T, typename U> ostream &operator<<(ostream &os, pair<T,U> p){os << "(" << p.fst << "," << p.snd << ")"; return os;}
template<typename T, typename U> istream &operator>>(istream &is, pair<T,U> &p){is >> p.fst >> p.snd; return is;}
template<typename T> istream &operator>>(istream &is, vector<T> &v){FOR(i, 0, v.size()) is >> v[i]; return is;}
template<typename T> ostream &operator<<(ostream &os, vector<T> v){for(T x : v) os << x << " "; return os;}
template<typename T> ostream &operator<<(ostream &os, set<T> s){for(T x : s) os << x << " "; return os;} 
//ostream &operator<<(ostream &os, ordered_set s){for(int x : s) os << x << " "; return os;} 
//ostream &operator<<(ostream &os, ordered_multiset s){for(pii x : s) os << x.fst << " "; return os;} 
template<typename T> ostream &operator<<(ostream &os, multiset<T> s){for(T x : s) os << x << " "; return os;} 
template<typename T, typename U> ostream &operator<<(ostream &os, map<T,U> m){for(auto x : m) os << x << " "; return os;} 
template<typename T> T sum(vector<T> v){T summ = 0; FOR(i, 0, sz(v)) summ += v[i]; return summ;}
ll mod_pow(ll a, ll b){ ll sol = 1; while(b){ if(b&1){ sol = (sol*a)%mod; b--; }else{ a = (a*a)%mod; b/=2; } } return sol;}
ll rem(ll a, ll b){ ll res = a%b; return res < 0 ? res+b : res; }
void test_case(){
    string s; cin >> s;
    int n = sz(s);
    int cnt = 0;
    for(char c : s) cnt += c == '1';
    int cons = 1;
    FOR(i, 1, n) cons += s[i] == '1' && s[i-1] == '1';
    if(cnt == 1){
        cout << "No\n";
    }else if(cons == cnt || cnt == 2){
        cout << "Yes\n";
    }else{
        cout << "No\n";
    }
}
int main(){
    #ifndef DEBUG
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    #endif
    int t = 1;
    FOR(i, 0, t) test_case();
    return 0;
}
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