Competitive Programming Solutions

arc107_a - Simple Math

24 Nov 2020 — Tags: math — URL

We only need to manipulate the summation a bit to represent it in closed forms. For example:

\[\sum_{i = 1}^{C} = \frac{C * (C + 1)}{2}\]

Since we’are dealing with modular arithmetic, we’ll need to compute the inverse of 2. The final formula is:

\[\frac{A * (A + 1)}{2} \times \frac{B * (B + 1)}{2} \times \frac{C * (C + 1)}{2}\]

Time complexity: $O(log(MOD))$

Memory complexity: $O(1)$

Click to show code.

 

using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
template <int M, typename T = long long>
class NumMod
{
    static_assert(std::is_integral<T>::value, "Integral required.");
    using NM = NumMod<M, T>;
    T x;
  public:
    static const int MOD = M;
    NumMod(T x) : x(x) {}
    NumMod() : x(0) {}
    NumMod(NM const &y) : x(y.v()) {}
    explicit operator T() const { return x; }
    T v(void) const { return (this->x + M) % M; }
    NM & operator=(NM const &y)
    {
        this->x = y.v();
        return *this;
    }
    NM &operator=(T const &y) { return this->operator=(NM(y)); }
    NM &operator+=(NM const &y) { return this->operator=(operator+(y)); }
    NM &operator-=(NM const &y) { return this->operator=(operator-(y)); }
    NM &operator*=(NM const &y) { return this->operator=(operator*(y)); }
    NM operator+(NM const &y) const { return (v() + y.v()) % M; }
    NM operator+(T const &y) const { return this->operator+(NM(y)); }
    NM operator-(NM const &y) const { return (v() - y.v()) % M; }
    NM operator-(T const &y) const { return this->operator-(NM(y)); }
    NM operator*(NM const &y) const { return (v() * y.v()) % M; }
    NM operator*(T const &y) const { return this->operator*(NM(y)); }
    NM operator/(NM const &y) const { return this->operator*(inverse(y)); }
};
ll const MOD = 998244353;
using NM = NumMod<MOD, ll>;
template <typename T>
T binpow(T a, ll b)
{
    T ans = 1;
    while (b > 0)
    {
        if (b & 1)
            ans *= a;
        a *= a;
        b >>= 1;
    }
    return ans;
}
NM inverse(NM const &y) { return binpow(y, y.MOD - 2); }
int main(void)
{
    ios::sync_with_stdio(false), cin.tie(NULL);
    ll a, b, c;
    cin >> a >> b >> c;
    NM ans = NM(a * (a + 1)) / 2 * NM(b * (b + 1)) / 2 * NM(c * (c + 1)) / 2;
    cout << ll(ans) << endl;
    return 0;
}

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