TopCoder SRM 660 Div2 Problem 1000 - Powerit (数论)

题意

求∑ni=1i2k−1modm

思路

求abmodc还有这么一个公式.

ab≡(amodc)bmodϕ(c)+ϕ(c)(modc),b>=ϕ(c)

因为m最大是1e9，那么k在30以内的我们都可以暴力得出结果，k>30就一定满足这个公式的条件，套上去算就行。

代码

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;
}
}
};