解题思路：

直接求C(n+m , m) % p , 由于n , m ,p都非常大，所以要用Lucas定理来解决大组合数取模的问题。

#include <string.h>

#include <iostream>

#include <algorithm>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <string>

#include <math.h>

#include <cstdio>

#include <stdlib.h>

#include <time.h>

#include <assert.h>

#define LL long long

#define FOR(i, x, y) for(int i=x;i<=y;i++)

using namespace std;

const int maxn = 100000 + 10;

LL Pow_mod(LL a, LL b, LL mod)

{

	LL ret = 1;

	while(b)

	{

		if(b & 1) ret = ret * a % mod;

		a = a * a % mod;

		b >>= 1;

	}

	return ret;

}

LL fac[maxn];

void init(LL p)

{

	fac[0] = 1;

	FOR(i, 1, p) fac[i] = (fac[i-1] * i) % p;

}

LL Lucas(LL n, LL m, LL p)

{

	LL ret = 1;

	while(n && m)

	{

		LL a = n % p, b = m % p;

		if(a < b) return 0;

		ret = (ret * fac[a] * Pow_mod(fac[b] * fac[a-b] % p , p - 2 , p)) % p;

		n /= p, m /= p;

	}

	return ret;

}

int main()

{

	int T;

	scanf("%d",&T);

	while(T--)

	{

		int n, m, p;

		scanf("%d%d%d", &n, &m, &p);

		init(p);

		printf("%d\n", (int)Lucas(n + m, m, p));

	}

	return 0;

}