​

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， 

#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;
}