组合数问题

1.递推处理O(n^2)

#include <bits/stdc++.h>
using namespace std;
const int N = 1003, null = INT64_MAX;
typedef long long ll;
ll m, x, y, p, n;
int c[N][N];
double eps=1e-6;
void init(){
	
	for(int i=0;i<N;i++){
		for(int j=0;j<=i;j++){
			if(!j)c[i][j]=1;
			else c[i][j]=(c[i-1][j-1]+c[i-1][j]);
		}
	}
}
int main(){

	int n;
	cin>>n;
		init();
	while(n--){
		int a,b;
		cin>>a>>b;
		cout<<c[a][b]<<endl;
	}
}

2.通过公式进行逆元运算N log(N)

#include <bits/stdc++.h>
using namespace std;
const int N = 1000003, null = INT64_MAX;
typedef long long ll;
ll fact[N],infact[N],mod=1e9+7;
ll m, x, y, p, n;

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

}

卢卡斯定理求组合数

#include <bits/stdc++.h>
using namespace std;
const int N = 1000003, null = INT64_MAX;
typedef long long ll;
ll fact[N],infact[N],mod=1e9+7;
ll  p;


ll qmi(ll a,ll b){
	ll res=1;
	while(b){
		if(b&1)res=res*a%p;
		a=a*a%p;
		b>>=1;
	}
   return res;
}
ll C(ll a,ll b){
	
    ll ans=1;
	for(int i=1,j=a;i<=b;i++,j--){
	   ans=ans*j%p;
	   ans=ans*qmi(i,p-2)%p;
	   	
	}
	
	return ans;
}

ll lucass(ll a,ll b){
	if(a<p&&b<p)
	return C(a,b);
	else return C(a%p,b%p)*lucass(a/p,b/p)%p;
}

int main(){
      int n;
      cin>>n;
      while(n--){
	  	int a,b;
	  	cin>>a>>b>>p;
	   cout<<lucass(a,b)%p<<endl;
	  }

}

不约分的大数排列组合

#include <bits/stdc++.h>
using namespace std;
const int N = 1000003, null = INT64_MAX;
typedef long long ll;
ll fact[N], infact[N], mod = 1e9 + 7;
bool f[N];
int prim[N], sum[N], cnt;
void sushu(int n) {
	for (int i = 2; i <= n; i++) {
		if (!f[i]) 
			prim[cnt++] = i;
		for (int j = 0; j <= n / i; j++) {
			f[i * prim[j]] = true;
			if (i % prim[j] == 0)break;
		}
	}

}
int get(int n, int p) {
	int res = 0;
	while (n) {
		res += n / p;
		n = n / p;
	}
	return res;
}
vector<int> mul(vector<int> a, int b) {
	vector<int> c;
	int t = 0;
	for (int i = 0; i < a.size(); i++) {
		t += a[i] * b;
		c.push_back(t % 10);
		t /= 10;
	}
	while (t) {
		c.push_back(t % 10);
		t /= 10;
	}
	return c;
}
int main() {
	int a, b;
	cin>>a>>b;
	sushu(a);
	for (int i = 0; i < cnt; i++) {
		int p=prim[i];
	    sum[i]=get(a,p)-get(b,p)-get(a-b,p);
	}
	  vector<int> v;
	  v.push_back(1);
	  for(int i=0;i<cnt;i++){
	  	for(int j=0;j<sum[i];j++){
		  	v=mul(v,prim[i]);
		  }
	  }
	  for(int i=v.size()-1;i>=0;i--)cout<<v[i];
}