组合数 2.1 2.2

 O(nlogn)预处理， O(1)查询

#include
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'

using namespace std;

typedef pair PII;
typedef long long ll;
typedef long double ld;

const int N = 100010, mod = 1e9 + 7;

int n;
int fact[N], infact[N];

int qmi(int a, int k)
{
	int res = 1;
	while(k)
	{
		if(k & 1)res = (ll)res * a % mod;
		a = (ll)a * a % mod;
		k >>= 1;
	}
	return res;
}

int C(int a, int b)
{
	return (ll)fact[a] * infact[b] % mod * infact[a - b] % mod;
}

int main()
{
	IOS
	
	cin >> n;
	fact[0] = infact[0] = 1;
	for(int i = 1; i < N; i ++)
	{
		fact[i] = ((ll)fact[i - 1] * i) % mod;
		infact[i] = qmi(fact[i], mod - 2);
	}
	
	while(n --)
	{
		int a, b;
		cin >> a >> b;
		cout << C(a, b) << endl;
	}
	
	return 0;
}

O(n)预处理，O(1)查询

#include
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'

using namespace std;

typedef pair PII;
typedef long long ll;
typedef long double ld;

const int N = 100010, mod = 1e9 + 7;

int n;
int fact[N], infact[N], inv[N];

int C(int a, int b)
{
	return (ll)fact[a] * infact[b] % mod * infact[a - b] % mod;
}

int main()
{
	IOS
	
	cin >> n;
	fact[0] = fact[1]= infact[0] = infact[1] = inv[1] = 1;
	for(int i = 2; i < N; i ++)
	{
		fact[i] = ((ll)fact[i - 1] * i) % mod;
		inv[i] = (ll)(mod - mod / i) * inv[mod % i] % mod;
		infact[i] = (ll)infact[i - 1] * inv[i] % mod;
	}
	
	while(n --)
	{
		int a, b;
		cin >> a >> b;
		cout << C(a, b) << endl;
	}
	
	return 0;
}