結果

問題 No.2419 MMA文字列2
ユーザー drken1215drken1215
提出日時 2023-08-12 13:48:04
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 64 ms / 2,000 ms
コード長 11,079 bytes
コンパイル時間 2,373 ms
コンパイル使用メモリ 215,916 KB
実行使用メモリ 45,952 KB
最終ジャッジ日時 2024-11-19 16:14:03
合計ジャッジ時間 4,157 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge5
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
5,248 KB
testcase_01 AC 2 ms
5,248 KB
testcase_02 AC 2 ms
5,248 KB
testcase_03 AC 2 ms
5,248 KB
testcase_04 AC 2 ms
5,248 KB
testcase_05 AC 2 ms
5,248 KB
testcase_06 AC 2 ms
5,248 KB
testcase_07 AC 2 ms
5,248 KB
testcase_08 AC 2 ms
5,248 KB
testcase_09 AC 2 ms
5,248 KB
testcase_10 AC 2 ms
5,248 KB
testcase_11 AC 2 ms
5,248 KB
testcase_12 AC 16 ms
12,928 KB
testcase_13 AC 3 ms
5,248 KB
testcase_14 AC 29 ms
21,632 KB
testcase_15 AC 19 ms
14,720 KB
testcase_16 AC 40 ms
22,400 KB
testcase_17 AC 11 ms
8,960 KB
testcase_18 AC 31 ms
23,296 KB
testcase_19 AC 33 ms
23,936 KB
testcase_20 AC 12 ms
9,856 KB
testcase_21 AC 9 ms
7,808 KB
testcase_22 AC 61 ms
45,952 KB
testcase_23 AC 61 ms
45,952 KB
testcase_24 AC 62 ms
45,952 KB
testcase_25 AC 61 ms
45,952 KB
testcase_26 AC 9 ms
8,448 KB
testcase_27 AC 61 ms
45,824 KB
testcase_28 AC 64 ms
45,952 KB
testcase_29 AC 63 ms
45,824 KB
testcase_30 AC 64 ms
45,952 KB
testcase_31 AC 64 ms
45,824 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;
using pint = pair<int, int>;
using pll = pair<long long, long long>;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

#define REP(i, n) for (long long i = 0; i < (long long)(n); ++i)
#define REP2(i, a, b) for (long long i = a; i < (long long)(b); ++i)
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P)
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, deque<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, vector<vector<T> > P)
{ for (int i = 0; i < P.size(); ++i) { s << endl << P[i]; } return s << endl; }
template<class T> ostream& operator << (ostream &s, set<T> P)
{ for(auto it : P) { s << "<" << it << "> "; } return s; }
template<class T> ostream& operator << (ostream &s, multiset<T> P)
{ for(auto it : P) { s << "<" << it << "> "; } return s; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P)
{ for(auto it : P) { s << "<" << it.first << "->" << it.second << "> "; } return s; }


/*///////////////////////////////////////////////////////*/
// Union-Find, modint, segtree, lazy segtree
/*///////////////////////////////////////////////////////*/

// 4-neighbor (or 8-neighbor)
const vector<int> dx = {1, 0, -1, 0, 1, -1, 1, -1};
const vector<int> dy = {0, 1, 0, -1, 1, 1, -1, -1};

// Union-Find
struct UnionFind {
    // core member
    vector<int> par;

    // constructor
    UnionFind() { }
    UnionFind(int n) : par(n, -1) { }
    void init(int n) { par.assign(n, -1); }
    
    // core methods
    int root(int x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }
    
    bool same(int x, int y) {
        return root(x) == root(y);
    }
    
    bool merge(int x, int y) {
        x = root(x), y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }
    
    int size(int x) {
        return -par[root(x)];
    }
    
    // debug
    friend ostream& operator << (ostream &s, UnionFind uf) {
        map<int, vector<int>> groups;
        for (int i = 0; i < uf.par.size(); ++i) {
            int r = uf.root(i);
            groups[r].push_back(i);
        }
        for (const auto &it : groups) {
            s << "group: ";
            for (auto v : it.second) s << v << " ";
            s << endl;
        }
        return s;
    }
};

// modint
template<int MOD> struct Fp {
    // inner value
    long long val;
    
    // constructor
    constexpr Fp() noexcept : val(0) { }
    constexpr Fp(long long v) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr long long get() const noexcept { return val; }
    constexpr int get_mod() const noexcept { return MOD; }
    
    // arithmetic operators
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp &r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp &r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp &r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp &r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp &r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp &r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp &r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp &r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b, swap(a, b);
            u -= t * v, swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp pow(long long n) const noexcept {
        Fp res(1), mul(*this);
        while (n > 0) {
            if (n & 1) res *= mul;
            mul *= mul;
            n >>= 1;
        }
        return res;
    }
    constexpr Fp inv() const noexcept {
        Fp res(1), div(*this);
        return res / div;
    }

    // other operators
    constexpr bool operator == (const Fp &r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp &r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr istream& operator >> (istream &is, Fp<MOD> &x) noexcept {
        is >> x.val;
        x.val %= MOD;
        if (x.val < 0) x.val += MOD;
        return is;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD> &x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &r, long long n) noexcept {
        return r.pow(n);
    }
    friend constexpr Fp<MOD> modinv(const Fp<MOD> &r) noexcept {
        return r.inv();
    }
};

// Binomial coefficient
template<class T> struct BiCoef {
    vector<T> fact_, inv_, finv_;
    constexpr BiCoef() {}
    constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
        init(n);
    }
    constexpr void init(int n) noexcept {
        fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
        int MOD = fact_[0].get_mod();
        for(int i = 2; i < n; i++){
            fact_[i] = fact_[i-1] * i;
            inv_[i] = -inv_[MOD%i] * (MOD/i);
            finv_[i] = finv_[i-1] * inv_[i];
        }
    }
    constexpr T com(int n, int k) const noexcept {
        if (n < k || n < 0 || k < 0) return 0;
        return fact_[n] * finv_[k] * finv_[n-k];
    }
    constexpr T fact(int n) const noexcept {
        if (n < 0) return 0;
        return fact_[n];
    }
    constexpr T inv(int n) const noexcept {
        if (n < 0) return 0;
        return inv_[n];
    }
    constexpr T finv(int n) const noexcept {
        if (n < 0) return 0;
        return finv_[n];
    }
};

// Segment Tree
template<class Monoid> struct SegTree {
    using Func = function<Monoid(Monoid, Monoid)>;

    // core member
    int SIZE;
    Func F;
    Monoid IDENTITY;
    
    // data
    int offset;
    vector<Monoid> dat;

    // constructor
    SegTree() {}
    SegTree(int n, const Func &f, const Monoid &identity)
    : SIZE(n), F(f), IDENTITY(identity) {
        offset = 1;
        while (offset < n) offset *= 2;
        dat.assign(offset * 2, IDENTITY);
    }
    void init(int n, const Func &f, const Monoid &identity) {
        SIZE = n;
        F = f;
        IDENTITY = identity;
        offset = 1;
        while (offset < n) offset *= 2;
        dat.assign(offset * 2, IDENTITY);
    }
    int size() const { return SIZE; }
    
    // set, a is 0-indexed //
    // build(): O(N)
    void set(int a, const Monoid &v) { dat[a + offset] = v; }
    void build() {
        for (int k = offset - 1; k > 0; --k)
            dat[k] = F(dat[k*2], dat[k*2+1]);
    }
    void build(const vector<Monoid> &vec) {
        for (int a = 0; a < vec.size() && a + offset < dat.size(); ++a)
            set(a, vec[a]);
        build();
    }
    
    // update a, a is 0-indexed, O(log N)
    void update(int a, const Monoid &v) {
        int k = a + offset;
        dat[k] = v;
        while (k >>= 1) dat[k] = F(dat[k*2], dat[k*2+1]);
    }
    
    // get [a, b), a and b are 0-indexed, O(log N)
    Monoid get(int a, int b) {
        Monoid vleft = IDENTITY, vright = IDENTITY;
        for (int left = a + offset, right = b + offset; left < right;
        left >>= 1, right >>= 1) {
            if (left & 1) vleft = F(vleft, dat[left++]);
            if (right & 1) vright = F(dat[--right], vright);
        }
        return F(vleft, vright);
    }
    Monoid get_all() { return dat[1]; }
    Monoid operator [] (int a) const { return dat[a + offset]; }
    
    // get max r that f(get(l, r)) = True (0-indexed), O(log N)
    // f(IDENTITY) need to be True
    int max_right(const function<bool(Monoid)> f, int l = 0) {
        if (l == SIZE) return SIZE;
        l += offset;
        Monoid sum = IDENTITY;
        do {
            while (l % 2 == 0) l >>= 1;
            if (!f(F(sum, dat[l]))) {
                while (l < offset) {
                    l = l * 2;
                    if (f(F(sum, dat[l]))) {
                        sum = F(sum, dat[l]);
                        ++l;
                    }
                }
                return l - offset;
            }
            sum = F(sum, dat[l]);
            ++l;
        } while ((l & -l) != l);  // stop if l = 2^e
        return SIZE;
    }

    // get min l that f(get(l, r)) = True (0-indexed), O(log N)
    // f(IDENTITY) need to be True
    int min_left(const function<bool(Monoid)> f, int r = -1) {
        if (r == 0) return 0;
        if (r == -1) r = SIZE;
        r += offset;
        Monoid sum = IDENTITY;
        do {
            --r;
            while (r > 1 && (r % 2)) r >>= 1;
            if (!f(F(dat[r], sum))) {
                while (r < offset) {
                    r = r * 2 + 1;
                    if (f(F(dat[r], sum))) {
                        sum = F(dat[r], sum);
                        --r;
                    }
                }
                return r + 1 - offset;
            }
            sum = F(dat[r], sum);
        } while ((r & -r) != r);
        return 0;
    }
    
    // debug
    friend ostream& operator << (ostream &s, const SegTree &seg) {
        for (int i = 0; i < seg.size(); ++i) {
            s << seg[i];
            if (i != seg.size()-1) s << " ";
        }
        return s;
    }
};


vector<long long> calc_divisor(long long n) {
    vector<long long> res;
    for (long long i = 1LL; i*i <= n; ++i) {
        if (n % i == 0) {
            res.push_back(i);
            long long j = n / i;
            if (j != i) res.push_back(j);
        }
    }
    sort(res.begin(), res.end());
    return res;
}


int main() {
    string S;
    cin >> S;
    int N = S.size();
    
    vector<vector<long long>> sum(26, vector<long long>(N+1, 0));
    for (int i = 0; i < N; ++i) {
        int c = S[i] - 'A';
        for (int j = 0; j < 26; ++j) {
            sum[j][i+1] = sum[j][i] + (j == c);
        }
    }
    
    long long res = 0;
    for (int i = 0; i < N; ++i) {
        int c = S[i] - 'A';
        long long left = sum[c][i];
        long long right = (N - i - 1) - (sum[c][N] - sum[c][i+1]);
        res += left * right;
    }
    cout << res << endl;
}
0