hdu3037SavingBeans【大组合数取模-Lucas定理+逆元取模】

Lucas定理

A、B是非负整数，p是质数。A B写成p进制：A=a[n]a[n-1]...a[0]，B=b[n]b[n-1]...b[0]。
则组合数C(A,B)与C(a[n],b[n])*C(a[n-1],b[n-1])*...*C(a[0],b[0])  mod p同余

即：Lucas(n,m,p)=C(n%p,m%p)*Lucas(n/p,m/p,p) 

//快速幂a^b % k

ll PowerMod(ll a, ll b, ll k) {
    ll tmp = a, ret = 1;
    while (b) {
        if (b & 1) ret = ret * tmp % k;
        tmp = tmp * tmp % k;
        b >>= 1;
    }
    return ret;
}

//求C(n, m)%p   p最大为10^5    n, m可以很大！

ll Lucas(ll n, ll m, ll p) {
    ll ret = 1;
    while (n && m) {
        ll nn = n%p, mm = m%p;
        if (nn         //fac[nn]为预处理的 fac[n] = n!%p 
        ret = ret*fac[nn]*PowerMod(fac[mm]*fac[nn-mm]%p, p-2, p)%p;
        n /= p;
        m /= p;
    }
    return ret;
}
//C(n, m) % p
Lucas(n, m, p);

AC代码：

#include 
#include 
#include 
#include 
using namespace std;

typedef long long ll;
ll fac[100003];

void init(ll p) {
    fac[0] = 1;
    for (int i=1; i<=p; i++) fac[i] = fac[i-1]*i%p;
}
ll PowerMod(ll a, ll b, ll k) {
    ll tmp = a, ret = 1;
    while (b) {
        if (b & 1) ret = ret * tmp % k;
        tmp = tmp * tmp % k;
        b >>= 1;
    }
    return ret;
}
ll Lucas(ll n, ll m, ll p) {
    ll ret = 1;
    while (n && m) {
        ll nn = n%p, mm = m%p;
        if (nn         ret = ret*fac[nn]*PowerMod(fac[mm]*fac[nn-mm]%p, p-2, p)%p;
        n /= p;
        m /= p;
    }
    return ret;
}

int main() {
    int T;
    ll n, m, p;
    cin >> T;
    while (T--) {
        cin >> n >> m >> p;
        init(p);
        cout <    }
    return 0;
}