一本通1651：【例 4】古代猪文

#include <cstdio>
#include <cmath>
#define ll long long
#define rll register ll
using namespace std;
const int mod = 999911658, su[5] = {0, 2, 3, 4679, 35617};
ll a[5], b[5], jc[5][36000], X, Y, n, g, ans, ans2 = 1;
inline ll pow(rll x, rll y, rll mo) {
    x %= mo;
    rll sum = 1;

    while (y > 0) {
        if (y & 1)
            sum = sum * x % mo;

        x = x * x % mo;
        y /= 2;
    }

    return sum;
}
inline ll zh(rll x, rll y, rll mo, rll u) {
    if (x < y)
        return 0;

    return jc[u][x] * pow(jc[u][y], mo - 2, mo) % mo * pow(jc[u][x - y], mo - 2, mo) % mo;
}
inline ll lcs(rll x, rll y, rll mo, rll u) {
    if (y == 0)
        return 1;

    return zh(x % mo, y % mo, mo, u) * lcs(x / mo, y / mo, mo, u) % mo;
}
inline ll crt(rll x) {
    rll sum = 0;

    for (rll i = 1; i < 5; i++) {
        rll y = lcs(n, x, su[i], i);
        sum = (sum + y * a[i] * b[i] % mod) % mod;
    }

    return (sum + mod) % mod;
}
inline void exgcd(rll w1, rll w2) {
    if (w2 == 0) {
        X = 1;
        Y = 0;
        return;
    }

    exgcd(w2, w1 % w2);
    rll U = X;
    X = Y;
    Y = U - w1 / w2 * Y;
}
int main() {
    for (rll i = 1; i < 5; i++) {
        a[i] = mod / su[i];
        exgcd(a[i], su[i]);
        b[i] = X;
        jc[i][0] = 1;

        for (rll j = 1; j <= su[i]; j++)
            jc[i][j] = jc[i][j - 1] * j % su[i];
    }

    scanf("%lld%lld", &n, &g);

    for (rll i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            ans = (ans + crt(i)) % mod;

            if (i != n / i && n % (n / i) == 0)
                ans = (ans + crt(n / i)) % mod;
        }
    }

    printf("%lld", pow(g, ans, mod + 1));
}