#794881 (C++23) No.2063 ±2^k operations (easy)

• 問題: No.2063 ±2^k operations (easy)
• ユーザー: k1suxu
• 提出日時: 2022-09-02 22:24:35
• 言語: C++23 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 5 ms / 2,000 ms
• コード長: 2,279 bytes
• コンパイル時間: 3,478 ms
• コンパイル使用メモリ: 251,268 KB
• 実行使用メモリ: 6,820 KB
• 最終ジャッジ日時: 2024-11-16 04:19:38
• 合計ジャッジ時間: 4,366 ms
• ジャッジサーバーID （参考情報）: judge1 / judge5

ファイルパターン	結果
sample	AC * 3
other	AC * 21

ソースコード

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#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
#define FOR(n) for(int i = 0; i < (int)n; i++)
#define repi(i,a,b) for(int i = (int)a; i < (int)b; i++)
#define pb push_back
#define m0(x) memset(x,0,sizeof(x))
#define fill(x,y) memset(x,y,sizeof(x))
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
//#define mp make_pair
#define vi vector<int>
#define vvi vector<vi>
#define vll vector<ll>
#define vvll vector<vll>
#define vs vector<string>
#define vvs vector<vs>
#define vc vector<char>
#define vvc vector<vc>
#define pii pair<int,int>
#define pllll pair<ll,ll>
#define vpii vector<pair<int,int>>
#define vpllll vector<pair<ll,ll>>
#define vpis vector<pair<int,string>>
#define vplls vector<pair<ll, string>>
#define vpsi vector<pair<string, int>>
#define vpsll vector<pair<string, ll>>
template<typename T>
void chmax(T &a, const T &b) {a = (a > b? a : b);}
template<typename T>
void chmin(T &a, const T &b) {a = (a < b? a : b);}
using ll = long long;
using ld = long double;
using ull = unsigned long long;
const ll INF = numeric_limits<long long>::max() / 2;
const ld pi = 3.1415926535897932384626433832795028;
const ll mod = 1e9 + 7;
int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
int dy[] = {0, -1, 0, 1, -1, 1, -1, 1};
#define int long long
vector<pair<char, int>> run_length_encoding(string s) {
    char need = s[0];
    int cnt = 0;
    vector<pair<char, int>> ret;
    for(char c : s) {
        if(c == need) {
            cnt++;
        }else {
            ret.emplace_back(need, cnt);
            need = c;
            cnt = 1;
        }
    }
    if(cnt != 0) ret.emplace_back(need, cnt);
    return ret;
}
void solve() {
    string n;
    cin >> n;
    int zero = (int)count(all(n), '0');
    int one = (int)count(all(n), '1');
    if(one <= 1) {
        cout << "No" << endl;
    }else if(one == 2) {
        cout << "Yes" << endl;
    }else {
        vector<pair<char, int>> p = run_length_encoding(n);
        int cnt = 0;
        for(auto e : p) if(e.first == '1') cnt++;
        if(cnt == 1) {
            cout << "Yes" << endl;
        }else {
            cout << "No" << endl;
        }
    }
}
signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}
 
 
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