解题思路：

直接求C(n+m , m) % p , 因为n , m ,p都很大，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;
}

HDU 3037 Saving Beans(Lucas定理的直接应用）