lucas定理

记一下结论：

当p为质数时，C(n, a) = C(n % p, a % p) * C(n / p, a / p) (mod p)

#include <cstdio>

typedef long long LL;
const int N = 100010;

LL inv[N], nn[N], invn[N];

inline void getinv(int p) {
    inv[1] = 1;
    for(int i = 2; i < p; i++) {
        inv[i] = (p - (p / i)) * inv[p % i] % p;
    }
    return;
}

LL C(LL a, LL b, LL c) {
    return nn[a] * invn[b] % c * invn[a - b] % c;
}
LL lucas(LL a, LL b, LL c) {
    if(!b) {
        return 1;
    }
    return C(a % c, b % c, c) * lucas(a / c, b / c, c) % c;
}

inline void solve() {
    LL n, m, p;
    scanf("%lld%lld%lld", &n, &m, &p);

    getinv(p);
    for(int i = 1; i <= p; i++) {
        invn[i] = invn[i - 1] * inv[i] % p;
        nn[i] = nn[i - 1] * i % p;
    }

    LL ans = lucas(m + n, n, p);
    printf("%lld
", ans);
    return;
}

int main() {
    nn[0] = invn[0] = 1;
    int T;
    scanf("%d", &T);
    while(T--) {
        solve();
    }

    return 0;
}

复杂度O(p + logn)

---

exlucas就是把模数拆了，然后CRT合并。

以后来填。