結果

問題 No.2021 Not A but B
ユーザー UMRgurashiUMRgurashi
提出日時 2023-06-22 10:05:59
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 7,024 bytes
コンパイル時間 6,079 ms
コンパイル使用メモリ 309,300 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-29 17:46:38
合計ジャッジ時間 6,923 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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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
5,248 KB
testcase_01 AC 1 ms
5,376 KB
testcase_02 AC 1 ms
5,376 KB
testcase_03 AC 2 ms
5,376 KB
testcase_04 WA -
testcase_05 AC 1 ms
5,376 KB
testcase_06 WA -
testcase_07 AC 2 ms
5,376 KB
testcase_08 AC 6 ms
5,868 KB
testcase_09 AC 5 ms
5,868 KB
testcase_10 AC 6 ms
5,864 KB
testcase_11 AC 5 ms
5,980 KB
testcase_12 AC 5 ms
5,980 KB
testcase_13 AC 5 ms
5,984 KB
testcase_14 AC 5 ms
5,872 KB
testcase_15 AC 6 ms
5,920 KB
testcase_16 AC 5 ms
5,736 KB
testcase_17 AC 5 ms
5,856 KB
testcase_18 AC 5 ms
5,872 KB
testcase_19 AC 5 ms
5,868 KB
testcase_20 AC 5 ms
5,996 KB
testcase_21 AC 2 ms
5,376 KB
testcase_22 AC 2 ms
5,376 KB
testcase_23 AC 2 ms
5,376 KB
testcase_24 AC 2 ms
5,376 KB
testcase_25 AC 2 ms
5,376 KB
testcase_26 AC 2 ms
5,376 KB
testcase_27 AC 6 ms
5,892 KB
testcase_28 AC 5 ms
5,848 KB
testcase_29 WA -
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 target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <cstdlib>
#include <atcoder/all>
using namespace atcoder;
#include <chrono>

#define int long long
#define double long double
#define stoi stoll
//#define endl "\n"
using std::abs;

using namespace std;
constexpr double PI = 3.14159265358979323846;
const  int INF = 1LL << 61;
const int dx[8] = { 0,1,0,-1,1,1,-1,-1 };
const int dy[8] = { 1,0,-1,0,1,-1,1,-1 };



#define rep(i,n) for(int i=0;i<n;++i)
#define REP(i,n) for(int i=1;i<=n;i++)
#define sREP(i,n) for(int i=1;i*i<=n;++i)
#define krep(i,k,n) for(int i=(k);i<n+k;i++)
#define Krep(i,k,n) for(int i=(k);i<n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define Rrep(i,n) for(int i=n;i>0;i--)
#define frep(i,n) for(auto &x:n)
#define LAST(x) x[x.size()-1]
#define ALL(x) (x).begin(),(x).end()
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x)
#define RUD(a,b) (((a)+(b)-1)/(b))
#define sum1_n(n) ((n)*(n+1)/2)
#define SUM1n2(n) (n*(2*n+1)*(n+1))/6
#define SUMkn(k,n) (SUM1n(n)-SUM1n(k-1))
#define SZ(x) ((int)(x).size())
#define PB push_back
#define Fi first
#define Se second
#define lower(vec, i) *lower_bound(ALL(vec), i)
#define upper(vec, i) *upper_bound(ALL(vec), i)
#define lower_count(vec, i) (int)(lower_bound(ALL(vec), i) - (vec).begin())
#define acc(vec) accumulate(ALL(vec),0LL)
template<class... T>
constexpr auto min(T... a) {
    return min(initializer_list<common_type_t<T...>>{a...});
}

template<class... T>
constexpr auto max(T... a) {
    return max(initializer_list<common_type_t<T...>>{a...});
}

template<class... T>
void in(T&... a) {
    (cin >> ... >> a);
}


void out() { cout << "\n"; }
template <typename T, class... U, char sep = ' '>
void out(const T& t, const U &...u) {
    cout << t;
    if (sizeof...(u)) cout << sep;
    out(u...);
}

template <typename T>
bool nxp(vector<T>& v) {
    return next_permutation(begin(v), end(v));
}

#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)

template <class T>
using v = vector<T>;
template <class T>
using vv = vector<v<T>>;
template <class T>
using vvv = vector<vv<T>>;

using pint = pair<int, int>;
using tint = tuple<int, int, int>;
using qint = tuple<int, int, int, int>;

double LOG(int a, int b) {
    return log(b) / log(a);
}
int DISTANCE(pint a, pint b) {
    return (abs(a.first - b.first) * abs(a.first - b.first) + abs(a.second - b.second) * abs(a.second - b.second));
}

inline bool BETWEEN(int x, int min, int max) {
    if (min <= x && x <= max)
        return true;
    else
        return false;
}
inline bool between(int x, int min, int max) {
    if (min < x && x < max) return true;
    else return false;
}
inline bool BETWEEN2(int i, int j, int H, int W) {
    if (BETWEEN(i, 0, H - 1) && BETWEEN(j, 0, W - 1)) return true;
    else return false;
}

template<class T>
inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T>
inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

inline bool bit(int x, int i) {
    return x >> i & 1;
}


void  yn(bool x) {
    if (x) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
}
void  YN(bool x) {
    if (x) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO" << endl;
    }
}

int ipow(int x, int n) {
    int ans = 1;
    while (n > 0) {
        if (n & 1) ans *= x;
        x *= x;
        n >>= 1;
    }
    return ans;
}

template <typename T>
vector<T> compress(vector<T>& X) {
    vector<T> vals = X;
    sort(ALL(vals));
    vals.erase(unique(ALL(vals)), vals.end());
    rep(i, SZ(X))
        X[i] = lower_bound(ALL(vals), X[i]) - vals.begin();
    return vals;
}

v<pint> prime_factorize(int N) {
    v<pint>  res;
    for (int i = 2; i * i <= N; i++) {
        if (N % i != 0) continue;
        int ex = 0;
        while (N % i == 0) {
            ++ex;
            N /= i;
        }
        res.push_back({ i, ex });
    }
    if (N != 1) res.push_back({ N, 1 });
    return res;
}



struct Eratosthenes {
    v<bool> isprime;
    v<int> minfactor;

    Eratosthenes(int N) : isprime(N + 1, true),
        minfactor(N + 1, -1) {
        isprime[0] = false;
        isprime[1] = false;
        minfactor[1] = 1;
        for (int p = 2; p <= N; ++p) {
            if (!isprime[p]) continue;
            minfactor[p] = p;
            for (int q = p * 2; q <= N; q += p) {
                isprime[q] = false;
                if (minfactor[q] == -1) minfactor[q] = p;
            }
        }
    }
    v<pint> factorize(int n) {
        v<pint> res;
        while (n > 1) {
            int p = minfactor[n];
            int exp = 0;
            while (minfactor[n] == p) {
                n /= p;
                ++exp;
            }
            res.emplace_back(p, exp);
        }
        return res;
    }
};

int number_of_divisors(v<pint> p) {
    int ans = 1;
    for (pint x : p) {
        ans *= x.second + 1;
    }
    return ans;
}
int sum_of_divisors(v<pint> p) {
    int ans = 1;
    for (pint x : p) {

    }
    return ans;
}

//constexpr int MOD = 1000000007;
constexpr int MOD = 998244353;
//using mint = modint1000000007;
//using mint = modint998244353;
//using mint = static_modint<16637>;
using mint = modint;

vector<int> prime_enumerate(int N) {
    vector<bool> sieve(N / 3 + 1, 1);
    for (int p = 5, d = 4, i = 1, sqn = sqrt(N); p <= sqn; p += d = 6 - d, i++) {
        if (!sieve[i]) continue;
        for (int q = p * p / 3, r = d * p / 3 + (d * p % 3 == 2), s = 2 * p,
            qe = sieve.size();
            q < qe; q += r = s - r)
            sieve[q] = 0;
    }
    vector<int> ret{ 2, 3 };
    for (int p = 5, d = 4, i = 1; p <= N; p += d = 6 - d, i++)
        if (sieve[i]) ret.push_back(p);
    while (!ret.empty() && ret.back() > N) ret.pop_back();
    return ret;
}

vector<pair<char, int>> run_length(string x) {
    //{char,個数}
    vector<pair<char, int>> ans;
    char ch = x[0];
    int cou = 1;
    REP(i, SZ(x)) {
        if (x[i] == x[i - 1]) {
            cou++;
        }
        else {
            ans.push_back({ ch,cou });
            ch = x[i];
            cou = 1;
        }
    }
    return ans;
}

void solve() {
    inl(N);
    ins(S);
    auto s = run_length(S);
    int ans = 0;
    bool b= false;
    rep(i,SZ(s)) {
        if (s[i].first == 'A') {
            if (s[i].second == 1)ans++;
            else {
                if (i == 0 || i == SZ(s) - 1)
                    ans += s[i].second;
                else
                    ans += s[i].second + 1;
            }
        }
        else {
            if (s[i].second > 1)b = true;
        }
    }
    if (b)ans++;
    cout << ans;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(14);
    //cout << setfill('0') << right << setw(3);
    solve();
}
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