hdu 3037 lucas定理

思路：求c（m+n，n）%p

lucas：

a=ak*p^k+(ak-1)*p^(k-1)+(ak-2)*p^(k-2)+…+a0

b=bk*p^k+(bk-1)*p^(k-1)+(bk-2)*p^(k-2)+…+b0

C(a,b)=C(ak,bk)*C((ak-1),(bk-1))*…C(a0,b0)(modp)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long m,n,p;
long long f[100005];
void init()
{
    f[0]=1;
    f[1]=1;
    for(int i=2;i<=p;i++)f[i]=(f[i-1]*i)%p;
}
long long quickpow(long long m,long long n,int k)
{
    long long b = 1;
    while (n > 0)
    {
        if (n & 1)
            b = (b*m)%k;
        n = n >> 1 ;
        m = (m*m)%k;
    }
    return b;
}
long long C(long long a,long long b)
{
    if(a<b)return 0;
    else return f[a]*quickpow(f[a-b]*f[b]%p,p-2,p)%p;
}
long long lucas(long long a,long long b)
{
    if(b==0)return 1;
    else return (lucas(a/p,b/p)*C(a%p,b%p))%p;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&n,&m,&p);
        init();
        int ans=lucas(m+n,n)%p;
        cout<<ans<<endl;
    }
    return 0;
}