题意

给出n个树和m个种子，求把这m个种子放到n棵树中有多少中方法，可以选择不放。

思路

因为有可以不放的选择，所以我们可以看成是多加了n棵空树，所以答案就是C(mn+m)

因为n和m都很大，p比较小，所以直接用lucas就可以了。

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define Lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int maxn = 1e5 + 10;
const double eps = 1e-8;
const double PI = acos(-1.0);
typedef pair<int, int> pii;
int p;
LL fact[maxn];

void init()
{
    fact[0] = 1;
    for (int i = 1; i <= p; i++)
        fact[i] = fact[i-1] * i % p;
}

LL q_pow(LL x, LL n)
{
    LL res = 1;
    LL temp = x % p;
    while (n)
    {
        if (n & 1) res = res * temp % p;
        temp = temp * temp % p;
        n >>= 1;
    }
    return res;
}

LL inv(LL x)
{
    return q_pow(x, p - 2);
}

LL C(LL n, LL m)
{
    if (n < m) return 0;
    return fact[n] * inv(fact[m] * fact[n-m]);
}

LL Lucas(LL n, LL m, LL p)
{
    if (m == 0) return 1;
    return (C(n % p, m % p) * Lucas(n / p, m / p, p)) % p;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    cin >> T;
    while (T--)
    {
        LL n, m;
        cin >> n >> m >> p;
        init();
        cout << Lucas(n + m, n, p) << endl;
    }
    return 0;
}