結果

問題 No.1292 パタパタ三角形
ユーザー heno239heno239
提出日時 2020-11-22 12:20:33
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 26 ms / 2,000 ms
コード長 4,991 bytes
コンパイル時間 1,850 ms
コンパイル使用メモリ 129,796 KB
実行使用メモリ 9,772 KB
最終ジャッジ日時 2024-07-23 16:26:16
合計ジャッジ時間 2,791 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge5
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 * 14
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 # 

#pragma GCC optimize("Ofast")
#pragma GCC target ("sse4")

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
#include<array>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
constexpr ll mod = 998244353;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acosl(-1.0);

ll mod_pow(ll x, ll n, ll m = mod) {
    ll res = 1;
    while (n) {
        if (n & 1)res = res * x % m;
        x = x * x % m; n >>= 1;
    }
    return res;
}
struct modint {
    ll n;
    modint() :n(0) { ; }
    modint(ll m) :n(m) {
        if (n >= mod)n %= mod;
        else if (n < 0)n = (n % mod + mod) % mod;
    }
    operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, ll n) {
    if (n == 0)return modint(1);
    modint res = (a * a) ^ (n / 2);
    if (n % 2)res = res * a;
    return res;
}

ll inv(ll a, ll p) {
    return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
    fact[0] = modint(1);
    for (int i = 0; i < max_n - 1; i++) {
        fact[i + 1] = fact[i] * modint(i + 1);
    }
    factinv[max_n - 1] = modint(1) / fact[max_n - 1];
    for (int i = max_n - 2; i >= 0; i--) {
        factinv[i] = factinv[i + 1] * modint(i + 1);
    }
}
modint comb(int a, int b) {
    if (a < 0 || b < 0 || a < b)return 0;
    return fact[a] * factinv[b] * factinv[a - b];
}



void solve() {
    string s; cin >> s;
    int cx = 0, cy = 0;
    auto nex = [&](int& x, int& y, char c) {
        int nx, ny;
        if (c == 'a') {
            if (x % 3 == 2) {
                nx = x + 1, ny = y;
            }
            else if (x % 3 == 0) {
                nx = x-1, ny = y;
            }
            else {
                if (x % 6 == 1) {
                    if (y % 2 == 0)nx = x, ny = y - 1;
                    else nx = x, ny = y + 1;
                }
                else {
                    if (y % 2 == 0)nx = x, ny = y + 1;
                    else nx = x, ny = y - 1;
                }
            }
        }
        else if (c == 'b') {
            if (x % 3 == 0) {
                nx = x + 1, ny = y;
            }
            else if (x % 3 == 1) {
                nx = x - 1, ny = y;
            }
            else {
                if (x % 6 == 5) {
                    if (y % 2 == 0)nx = x, ny = y - 1;
                    else nx = x, ny = y + 1;
                }
                else {
                    if (y % 2 == 0)nx = x, ny = y + 1;
                    else nx = x, ny = y - 1;
                }
            }
        }
        else {
            if (x % 3 == 1) {
                nx = x + 1, ny = y;
            }
            else if (x % 3 == 2) {
                nx = x - 1, ny = y;
            }
            else {
                if (x % 6 == 3) {
                    if (y % 2 == 0)nx = x, ny = y - 1;
                    else nx = x, ny = y + 1;
                }
                else {
                    if (y % 2 == 0)nx = x, ny = y + 1;
                    else nx = x, ny = y - 1;
                }
            }
        }
        swap(x, nx); swap(y, ny);
    };
    vector<P> vs;
    cx = 500000, cy = 500001;
    vs.push_back({ cx,cy });
    rep(i, s.size()) {
        nex(cx, cy, s[i]);
        //cout << cx << " " << cy << "\n";
        vs.push_back({ cx,cy });
    }
    sort(all(vs));
    vs.erase(unique(all(vs)), vs.end());
    cout << vs.size() << "\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    //cout << fixed << setprecision(15);
    //init_f();
    //init();
    //expr();
    //int t; cin >> t; rep(i, t)
    solve();
    return 0;
}
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