题目大意：求
\[G^{\sum\limits_{d|N}\binom{n}{k}} mod\ \ 999911659
\]
\]
题解：卢卡斯定理+中国剩余定理
利用卢卡斯定理求出指数和式对各个素模数的解，再利用中国剩余定理合并四个解即可。
也可以在枚举 N 的因子的过程中，对于计算的四个解直接进行中国剩余定理的合并，答案不变。
代码如下
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 999911658;
const LL md[] = {2, 3, 4679, 35617};
const int maxn = 40000;
LL fac[maxn];
inline LL fpow(LL a, LL b, LL c) {
LL ret = 1 % c;
for (; b; b >>= 1, a = a * a % c) {
if (b & 1) {
ret = ret * a % c;
}
}
return ret;
}
inline LL comb(LL x, LL y, LL p) {
if (y > x) {
return 0;
}
return fac[x] * fpow(fac[x - y], p - 2, p) % p * fpow(fac[y], p - 2, p) % p;
}
LL Lucas(LL x, LL y, LL p) {
if (y == 0) {
return 1;
}
return Lucas(x / p, y / p, p) * comb(x % p, y % p, p) % p;
}
LL CRT(vector<LL> &v) {
LL ret = 0;
for (int i = 0; i < 4; i++) {
ret = (ret + mod / md[i] * fpow(mod / md[i], md[i] - 2, md[i]) % mod * v[i] % mod) % mod;
}
return ret;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
LL n, G;
cin >> n >> G;
if (G % (mod + 1) == 0) {
cout << 0 << endl;
return 0;
}
vector<LL> v;
for (int i = 0; i < 4; i++) {
LL res = 0;
fac[0] = 1;
for (int j = 1; j < 35617; j++) {
fac[j] = fac[j - 1] * j % md[i];
}
for (int j = 1; j <= sqrt(n); j++) {
if (n % j == 0) {
res = (res + Lucas(n, n / j, md[i])) % md[i];
if (j * j != n) {
res = (res + Lucas(n, j, md[i])) % md[i];
}
}
}
v.push_back(res);
}
LL p = CRT(v);
cout << fpow(G, p, mod + 1) << endl;
return 0;
}
/*
2
3
4679
35617
*/