题目大意：n个数，和不大于m的情况，结果模掉p，p保证为素数。
解题思路：隔板法，C(nn+m)多选的一块保证了n个数的和小于等于m。可是n，m非常大，所以用到Lucas定理。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
ll n, m, p;
ll qPow (ll a, ll k) {
ll ans = 1;
while (k) {
if (k&1)
ans = (ans * a) % p;
a = (a * a) % p;
k /= 2;
}
return ans;
}
/*
long long qPow(long long a, long long k) {
if (k == 0) return 1;
if (k == 1) return a;
long long ans = qPow(a * a % p, k>>1);
if (k&1) ans = ans * a % p;
return ans;
}
*/
ll C (ll a, ll b) {
if (a < b)
return 0;
if (b > a - b)
b = a - b;
ll up = 1, down = 1;
for (ll i = 0; i < b; i++) {
up = up * (a-i) % p;
down = down * (i+1) % p;
}
return up * qPow(down, p-2) % p;
}
ll lucas (ll a, ll b, ll p) {
if (b == 0)
return 1;
return C(a%p, b%p) * lucas(a/p, b/p, p) % p;
}
int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
scanf("%lld%lld%lld", &n, &m, &p);
printf("%lld\n", lucas(n+m, m, p));
}
return 0;
}