TopCoder SRM 660 Div2 Problem 1000 - Powerit (數論)

題意

求∑ni=1i2k−1modm

思路

求abmodc 還有這麼一個公式.

ab≡(amodc)bmodϕ(c)+ϕ(c)(modc),b>=ϕ(c)

因爲m最大是1e9，那麼k在30以內的我們都可以暴力得出結果，k>30就一定滿足這個公式的條件，套上去算就行。

代碼

class Powerit {
public:
    int get_phi(int m)
    {
        int k = sqrt(m+.5), ans = m;
        for (int i = 2; i <= k; i++) if (m % i == 0)
        {
            ans = ans / i * (i-1);
            while (m % i == 0) m /= i;
        }
        if (m > 1) ans = ans / m * (m-1);
        return ans;
    }

    LL pow_mod(LL a, LL m, LL n)
    {
        LL ret = 1;
        while (m)
        {
            if (m & 1) ret = ret * a % n;
            a = a * a % n;
            m >>= 1;
        }
        return ret;
    }

    int calc(int n, int k, int m) {
        LL ans = 0;
        if (k <= 30)
        {
            for (int i = 1; i <= n; i++) ans = (ans + pow_mod(i, (1<<k)-1, m)) % m;
            return ans;
        }
        else
        {
            int phi_m = get_phi(m);
            for (int i = 1; i <= n; i++) ans = (ans + pow_mod(i%m, (pow_mod(2, k, phi_m)-1 + phi_m) % phi_m + phi_m, m)) % m;
            return ans;
        }
    }
};