組合數模板

C(m, n)

<span style="font-size:18px;">const int mod=1e9+7;
typedef long long ll;
//返回d=gcd(a,b);和對應於等式ax+by=d中的x,y
ll extend_gcd(ll a,ll b,ll &x,ll &y)
{
    if(a==0&&b==0) return -1;//無最大公約數
    if(b==0){x=1;y=0;return a;}
    ll d=extend_gcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
//*********求逆元素*******************
//ax = 1(mod n)
ll mod_reverse(ll a,ll n)
{
    ll x,y;
    ll d=extend_gcd(a,n,x,y);
    if(d==1) return (x%n+n)%n;
    else return -1;
}

ll c(ll m,ll n)
{
	ll i,j,t1,t2,ans;
	t1=t2=1;
	for(i=n;i>=n-m+1;i--) t1=t1*i%mod;
	for(i=1;i<=m;i++) t2=t2*i%mod;
	return  t1*mod_reverse(t2,mod)%mod;
}
</span>

 

Lucas定理求組合數C(n, m)

 

#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
using namespace std;  
typedef long long LL;  
LL Power_mod(LL a, LL b, LL p)  
{  
    LL res = 1;  
    while(b!=0)  
    {  
        if(b&1) res = (res*a)%p;  
        a = (a*a)%p;  
        b >>= 1;  
    }  
    return res;  
}  
LL Comb(LL a,LL b, LL p)  
{  
    if(a < b) return 0;  
    if(a == b) return 1;  
    if(b > a-b) b = a-b;  
    LL ans = 1, ca = 1, cb = 1;  
    for(LL i=0; i<b; ++i)  
    {  
        ca = (ca*(a-i))%p;  
        cb = (cb*(b-i))%p;  
    }  
    ans = (ca*Power_mod(cb, p-2, p))%p;  
    return ans;  
}  
LL Lucas(int n, int m, int p)  
{  
    LL ans = 1;  
    while(n && m && ans)  
    {  
        ans = (ans * Comb(n%p, m%p, p))%p;  
        n /= p;  
        m /= p;  
    }  
    return ans;  
}  
int main()  
{  
    int n,m,p;  
    while(scanf("%d%d%d",&n,&m,&p) !=EOF)  
    {  
        printf("%lld\n",Lucas(n,m,p));  
    }  
    return 0;  
}