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。

#include 
     
      #include 
      
       #include 
       
        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;}