Light OJ 1102 problem makes problem combination number

Source: Light OJ 1102 problem makes problem

Test instructions: An integer n is decomposed into the number of K to add how many scheme numbers can be duplicated

Idea: M apples put n boxes how many schemes allow boxes empty boxes Null correspondence 0 answer is C (n+m-1, n-1)

First of all, if the answer is not allowed is C (m-1, n-1) intervening spaces m Apple has a m-1 gap in this m-1 neutral selection n-1 a divided into n parts each part is not empty

Now allowed to empty then add an apple to each box is equivalent to n+m an Apple into n parts n+m-1 a neutral n-1 C (n+m-1, n-1)

Each part is at least 1 to 1 is actually empty I started adding it to facilitate the use of the intervening spaces method

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
Const LL mod = 1000000007;
const int MAXN = 2000010;
return a^p mod n fast power
LL A[MAXN];
ll Pow_mod (ll A, ll P, ll N)
{
	ll ans = 1;
	while (p)
	{
		if (p&1)
		{
			ans *= A;
			Ans%= n;
		}
		A *= A;
		a%= n;
		P >>= 1;
	}
	return ans;
}

ll C (ll N, ll m)
{
	ll x = a[n], y = a[n-m]*a[m]%mod;
	return X*pow_mod (y, mod-2, MoD)%mod;
}
int main ()
{
	a[0] = 1;
	for (int i = 1; I <= 2000000; i++)
	{
		a[i] = a[i-1]*i;
		A[i]%= mod;
	}	
	int cas = 0; 
	int T;
	scanf ("%d", &t); 
	while (t--)
	{
		LL n, m;
		scanf ("%lld%lld", &n, &m);
		printf ("Case%d:%lld\n", ++cas, C (n+m-1, m-1));
	}
	return 0;
}