lgP3807 P为1E6以内质数的组合数

有T次询问，每次给出整数n,m,p，计算C(n+m,n)%p的值。输入保证p为质数。
1<=n,m,p<=1E5; 1<=T<=10

n很大，p为百万级以内的质数，并且需要多次求组合数时，一般用lucas定理来计算组合数：lucas(n,k,p) = lucas(n/p,k/p,p) * C(n%p,k%p,p)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,a,b) for(int i=a; i<=b; i++)
#define per(i,a,b) for(int i=b; i>=a; i--)

const int Z = 100000;
int fac[Z+1], ifac[Z+1];
int power(int a, int b, int p) {
    int r = 1;
    for (int t = a; b; b /= 2) {
        if (b & 1) r = r * t % p;
        t = t * t % p;
    }
    return r % p;
}
int comb(int n, int k, int p) {
    if (k > n) return 0;
    return fac[n] * ifac[k] % p * ifac[n-k] % p;
}
int lucas(int n, int k, int p) {
    if (k == 0) return 1;
    return lucas(n/p, k/p, p) * comb(n%p, k%p, p) % p;
}
void solve() {
    int n, m, p;
    cin >> n >> m >> p;
    fac[0] = 1;
    rep(i,1,p-1) fac[i] = fac[i-1] * i % p;
    ifac[p-1] = power(fac[p-1], p-2, p);
    per(i,0,p-2) ifac[i] = ifac[i+1] * (i+1) % p;
    cout << lucas(n+m, n, p) << "\n";
}

signed main() {
    cin.tie(0)->sync_with_stdio(0);
    int t = 1; cin >> t;
    while (t--) solve();
    return 0;
}